Class 9
Mathematics
Federal Board of Intermediate and Secondary Education (FBISE) (FBISE)
📚 Model Textbook of Mathematics Grade 9 (FBISE / NBF)
Mastery Guide: Geometry of Straight Lines - Inclination, Slope, 6 Standard Forms, Intersecting Angles & Real-World Modeling
📖 Chapter 8: Geometry of Straight Lines
📅 Updated: Sep 19, 2026
FBISE Class 9 Mathematics • Chapter 8
Mastery Guide: Geometry of Straight Lines
Single National Curriculum (SNC) • Inclination, Gradient (Slope), 6 Standard Forms of Equations, Angles Between Lines, Family of Lines & Real-World Modeling
📖 1. Unit Overview & Target Learning Outcomes
A straight line is one of the foundational geometric objects. In coordinate geometry, a straight line represents a continuous linear relationship between two variables $x$ and $y$. This chapter explores the algebraic characteristics of lines—measuring their steepness (slope/gradient), deriving the 6 classical standard equations of lines, determining intersections and angles between coplanar lines, exploring pencils (families) of lines, and applying linear equations to solve real-world rate problems.
🎯 Core Learning Outcomes:
- Inclination & Gradient (Slope): Define inclination $\theta \in [0^\circ, 180^\circ)$ and slope $m = \tan \theta = \frac{y_2 - y_1}{x_2 - x_1}$. Understand horizontal ($m=0$), vertical ($m=\text{undefined}$), positive ($0^\circ < \theta < 90^\circ$), and negative ($90^\circ < \theta < 180^\circ$) slopes.
- Parallel & Perpendicular Conditions: Prove that non-vertical lines are parallel iff $m_1 = m_2$, and perpendicular iff $m_1 \cdot m_2 = -1$ ($m_2 = -\frac{1}{m_1}$).
- Six Standard Forms of Straight Line Equations:
- Slope-Intercept Form: $y = mx + c$
- Point-Slope Form: $y - y_1 = m(x - x_1)$
- Two-Point Form: $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$ or $\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$
- Two-Intercept Form: $\frac{x}{a} + \frac{y}{b} = 1$
- Symmetric Form: $\frac{x - x_1}{\cos \alpha} = \frac{y - y_1}{\sin \alpha} = r$
- Normal Form: $x \cos \alpha + y \sin \alpha = p$ (where $p > 0$)
- General Linear Equation & Reductions: Transform $Ax + By + C = 0$ into each of the 6 standard forms; extract slope $m = -\frac{A}{B}$, intercepts $a = -\frac{C}{A}, b = -\frac{C}{B}$, and normal parameters $p = \frac{|C|}{\sqrt{A^2 + B^2}}$.
- Angle Between Intersecting Lines: Calculate acute/counter-clockwise angle $\theta$ from line $l_1$ to $l_2$ using $\tan \theta = \frac{m_2 - m_1}{1 + m_1 m_2}$.
- Point of Intersection & Family of Lines: Find simultaneous solutions of two lines and formulate the equation of a family of lines passing through their intersection: $L_1 + k L_2 = 0$.
- Real-World Linear Modeling: Model costs, speeds, hotel rentals, taxi fares, electricity tariffs, temperature scales ($F = \frac{9}{5}C + 32$), and linear trends.
💡 2. Kid-Friendly Tips for Success & Memory Hooks
🏃 "Rise Over Run"
Slope is vertical change ($\Delta y$) divided by horizontal change ($\Delta x$). You must RISE before you can RUN! If you run uphill from left to right, slope is positive (+); downhill is negative (−).
⚡ Parallel = Equal; Perp = Flip & Negate
Parallel lines never meet $\implies m_1 = m_2$. Perpendicular lines meet at $90^\circ \implies$ flip the fraction upside down and switch the sign: $m_2 = -\frac{1}{m_1}$ (Negative Reciprocal!).
🎯 Normal Form $p$ is Always Positive
In $x \cos \alpha + y \sin \alpha = p$, $p$ is the physical perpendicular distance from the origin $(0,0)$ to the line. Distance cannot be negative, so $p \ge 0$ always!
✨ Family of Lines Shortcut
Never solve for intersection first if you can use $L_1 + k L_2 = 0$. Just substitute the given point or slope to find $k$ in one clean step!
🌍 3. Real-World Connections & Applications
🚗 Fixed vs Variable Cost (Taxis & Utilities)
A taxi charging Rs. 1500 flag-drop fee plus Rs. 450 per 30 mins behaves exactly as $y = mx + c$, where $c$ is the fixed intercept and $m$ is the variable hourly rate.
🌡️ Thermometry & Sensor Calibration
Converting Celsius to Fahrenheit ($F = 1.8 C + 32$) uses two known points: freezing $(0, 32)$ and boiling $(100, 212)$ in Two-Point Form!
🚢 Maritime & Flight Navigation
Ships steering along latitude/longitude lines from Karachi calculate flight bearings using the slope $m = \frac{\Delta \text{Lat}}{\Delta \text{Long}}$ to arrive accurately.
🔑 4. Study Cues & Conceptual Inquiries
- Why does a vertical line have an undefined slope? Because run $\Delta x = 0$, leading to division by zero ($\frac{\Delta y}{0}$), and $\tan 90^\circ = \infty$.
- What is the geometric meaning of $m_1 m_2 = -1$? It represents two directions meeting at right angles ($90^\circ$). Because $\tan(\theta + 90^\circ) = -\cot \theta = -\frac{1}{\tan \theta}$, their slopes multiply to $-1$.
- How do you distinguish between $x$-intercept and $y$-intercept? The $x$-intercept $(a, 0)$ is where the line crosses the x-axis ($y=0$); the $y$-intercept $(0, b)$ is where it crosses the y-axis ($x=0$).
- What happens when $k$ varies in $L_1 + k L_2 = 0$? As $k$ takes any real number, the equation generates every straight line passing through the unique point of intersection of $L_1$ and $L_2$ (forming a "pencil" of lines).
🌟 5. In-Depth Concepts & Visual Architectural Models
A. Inclination ($\theta$) & Gradient ($m$) of a Straight Line
The inclination $\theta$ is the angle made by the line with the positive direction of the x-axis, measured in the anti-clockwise direction ($0^\circ \le \theta < 180^\circ$). The gradient (slope) $m$ is defined as the tangent of inclination:
$$m = \tan \theta = \frac{\text{Rise}}{\text{Run}} = \frac{y_2 - y_1}{x_2 - x_1}$$
Figure 8.1: Geometric Derivation of Gradient $m = \tan \theta = \frac{y_2 - y_1}{x_2 - x_1}$
B. Master Matrix: The 6 Standard Forms of Straight Lines
| Form Name | Standard Equation | Given Parameters | Key Characteristics |
|---|---|---|---|
| 1. Slope-Intercept | $y = mx + c$ | Slope $m$, y-intercept $c$ | Passes through origin if $c=0$. |
| 2. Point-Slope | $y - y_1 = m(x - x_1)$ | Point $(x_1, y_1)$, Slope $m$ | Most versatile working form. |
| 3. Two-Point | $\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$ | Two points $(x_1, y_1), (x_2, y_2)$ | $m = \frac{y_2 - y_1}{x_2 - x_1}$ directly embedded. |
| 4. Two-Intercept | $\frac{x}{a} + \frac{y}{b} = 1$ | x-intercept $a$, y-intercept $b$ | Area of triangle with axes $= \frac{1}{2}|ab|$. |
| 5. Symmetric | $\frac{x - x_1}{\cos \alpha} = \frac{y - y_1}{\sin \alpha} = r$ | Point $(x_1, y_1)$, Inclination $\alpha$ | $r$ is directed distance along line. |
| 6. Normal Form | $x \cos \alpha + y \sin \alpha = p$ | Perp length $p > 0$, normal angle $\alpha$ | Derived from perpendicular from origin. |
Figure 8.2: Normal Form Geometry showing Perpendicular $p$ and Normal Angle $\alpha$
C. Angle Between Two Lines & Family of Lines
If $l_1$ and $l_2$ have slopes $m_1$ and $m_2$, the counter-clockwise angle $\theta$ from $l_1$ to $l_2$ is given by:
$$\tan \theta = \frac{m_2 - m_1}{1 + m_1 m_2}$$
For any two intersecting lines $L_1: a_1 x + b_1 y + c_1 = 0$ and $L_2: a_2 x + b_2 y + c_2 = 0$, every straight line passing through their point of intersection is represented by the Family of Lines:
$$(a_1 x + b_1 y + c_1) + k (a_2 x + b_2 y + c_2) = 0, \quad k \in \mathbb{R}$$
📝 6. Solved Textbook Exercises (Complete Step-by-Step Manual)
Exercise 8.1 • Gradient, Inclination, Parallel & Perpendicular Lines, Collinearity & Triangles
Q1. Find the gradient (slope) of line whose inclination is:
(i) $0^\circ$: $m = \tan 0^\circ = \mathbf{0}$
(ii) $30^\circ$: $m = \tan 30^\circ = \frac{1}{\sqrt{3}} \approx \mathbf{0.577}$
(iii) $60^\circ$: $m = \tan 60^\circ = \sqrt{3} \approx \mathbf{1.732}$
(iv) $90^\circ$: $m = \tan 90^\circ = \mathbf{\text{Undefined}}$ (Vertical line)
(v) $120^\circ$: $m = \tan 120^\circ = \tan(180^\circ - 60^\circ) = -\tan 60^\circ = -\sqrt{3} \approx \mathbf{-1.732}$
(vi) $150^\circ$: $m = \tan 150^\circ = -\tan 30^\circ = -\frac{1}{\sqrt{3}} \approx \mathbf{-0.577}$
(vii) $170^\circ$: $m = \tan 170^\circ = -\tan 10^\circ \approx \mathbf{-0.176}$
(viii) $45.5^\circ$: $m = \tan 45.5^\circ \approx \mathbf{1.0176}$
(ii) $30^\circ$: $m = \tan 30^\circ = \frac{1}{\sqrt{3}} \approx \mathbf{0.577}$
(iii) $60^\circ$: $m = \tan 60^\circ = \sqrt{3} \approx \mathbf{1.732}$
(iv) $90^\circ$: $m = \tan 90^\circ = \mathbf{\text{Undefined}}$ (Vertical line)
(v) $120^\circ$: $m = \tan 120^\circ = \tan(180^\circ - 60^\circ) = -\tan 60^\circ = -\sqrt{3} \approx \mathbf{-1.732}$
(vi) $150^\circ$: $m = \tan 150^\circ = -\tan 30^\circ = -\frac{1}{\sqrt{3}} \approx \mathbf{-0.577}$
(vii) $170^\circ$: $m = \tan 170^\circ = -\tan 10^\circ \approx \mathbf{-0.176}$
(viii) $45.5^\circ$: $m = \tan 45.5^\circ \approx \mathbf{1.0176}$
Q2. Find inclination of the line whose slope is:
(i) $0$: $\theta = \tan^{-1}(0) = \mathbf{0^\circ}$
(ii) $0.577$: $\theta = \tan^{-1}(0.577) = \mathbf{30^\circ}$
(iii) $-1.732$: $\theta = 180^\circ - \tan^{-1}(1.732) = 180^\circ - 60^\circ = \mathbf{120^\circ}$
(iv) $-0.364$: $\theta = 180^\circ - \tan^{-1}(0.364) = 180^\circ - 20^\circ = \mathbf{160^\circ}$
(ii) $0.577$: $\theta = \tan^{-1}(0.577) = \mathbf{30^\circ}$
(iii) $-1.732$: $\theta = 180^\circ - \tan^{-1}(1.732) = 180^\circ - 60^\circ = \mathbf{120^\circ}$
(iv) $-0.364$: $\theta = 180^\circ - \tan^{-1}(0.364) = 180^\circ - 20^\circ = \mathbf{160^\circ}$
Q3. Find gradient and inclination of lines joining:
(i) $A(2, 6), B(5, 8)$: $m = \frac{8-6}{5-2} = \mathbf{\frac{2}{3}} \approx 0.667$; $\theta = \tan^{-1}(0.667) \approx \mathbf{33.69^\circ}$
(ii) $C(-2, 4), D(1, -3)$: $m = \frac{-3-4}{1 - (-2)} = \mathbf{-\frac{7}{3}} \approx -2.333$; $\theta = 180^\circ - \tan^{-1}(2.333) \approx \mathbf{113.2^\circ}$
(iii) $E(5, -2), F(-2, -3)$: $m = \frac{-3 - (-2)}{-2 - 5} = \frac{-1}{-7} = \mathbf{\frac{1}{7}} \approx 0.143$; $\theta = \tan^{-1}(0.143) \approx \mathbf{8.13^\circ}$
(ii) $C(-2, 4), D(1, -3)$: $m = \frac{-3-4}{1 - (-2)} = \mathbf{-\frac{7}{3}} \approx -2.333$; $\theta = 180^\circ - \tan^{-1}(2.333) \approx \mathbf{113.2^\circ}$
(iii) $E(5, -2), F(-2, -3)$: $m = \frac{-3 - (-2)}{-2 - 5} = \frac{-1}{-7} = \mathbf{\frac{1}{7}} \approx 0.143$; $\theta = \tan^{-1}(0.143) \approx \mathbf{8.13^\circ}$
Q4. If $A(-2, 6)$ and $B(7, -3)$, find the slope of line:
Slope of $AB = m = \frac{-3-6}{7 - (-2)} = \frac{-9}{9} = -1$.
(i) Parallel to $AB$: $m_{\parallel} = m = \mathbf{-1}$
(ii) Perpendicular to $AB$: $m_{\perp} = -\frac{1}{m} = -\frac{1}{-1} = \mathbf{1}$
(i) Parallel to $AB$: $m_{\parallel} = m = \mathbf{-1}$
(ii) Perpendicular to $AB$: $m_{\perp} = -\frac{1}{m} = -\frac{1}{-1} = \mathbf{1}$
Q5. Find $x$ if the slope of line passing through $A(3, x), B(5, 8)$ is $4$:
$m = \frac{8-x}{5-3} = 4 \implies \frac{8-x}{2} = 4 \implies 8 - x = 8 \implies \mathbf{x = 0}$.
Q6. Find $k$ if lines through $A(k, 2), B(3, 5)$ and $C(5, -1), D(8, 7)$ are parallel:
$m_1 = \frac{5-2}{3-k} = \frac{3}{3-k}$; $m_2 = \frac{7 - (-1)}{8-5} = \frac{8}{3}$.
Parallel $\implies m_1 = m_2 \implies \frac{3}{3-k} = \frac{8}{3} \implies 9 = 8(3-k) = 24 - 8k \implies 8k = 15 \implies \mathbf{k = \frac{15}{8}}$.
Parallel $\implies m_1 = m_2 \implies \frac{3}{3-k} = \frac{8}{3} \implies 9 = 8(3-k) = 24 - 8k \implies 8k = 15 \implies \mathbf{k = \frac{15}{8}}$.
Q7. Find $k$ if lines through $P(-1, 2), Q(4, 7)$ and $R(2, k), S(7, 10)$ are perpendicular:
$m_1 = \frac{7-2}{4 - (-1)} = \frac{5}{5} = 1$; $m_2 = \frac{10-k}{7-2} = \frac{10-k}{5}$.
Perpendicular $\implies m_1 \cdot m_2 = -1 \implies 1 \cdot \left(\frac{10-k}{5}\right) = -1 \implies 10 - k = -5 \implies \mathbf{k = 15}$.
Perpendicular $\implies m_1 \cdot m_2 = -1 \implies 1 \cdot \left(\frac{10-k}{5}\right) = -1 \implies 10 - k = -5 \implies \mathbf{k = 15}$.
Q8. Using slopes, prove that $X(0, -3), Y(4, 7), Z(6, 12)$ are collinear:
Slope of $XY = \frac{7 - (-3)}{4-0} = \frac{10}{4} = \frac{5}{2}$.
Slope of $YZ = \frac{12-7}{6-4} = \frac{5}{2}$.
Since Slope($XY$) = Slope($YZ$) and $Y$ is a common point, points $X, Y, Z$ are collinear (Proved).
Slope of $YZ = \frac{12-7}{6-4} = \frac{5}{2}$.
Since Slope($XY$) = Slope($YZ$) and $Y$ is a common point, points $X, Y, Z$ are collinear (Proved).
Q9. Find $y$ if $P(4, y), Q(5, 2), R(6, 2y + 1)$ are collinear:
Slope of $PQ = \frac{2-y}{5-4} = 2-y$; Slope of $QR = \frac{(2y+1)-2}{6-5} = 2y-1$.
Collinear $\implies 2 - y = 2y - 1 \implies 3y = 3 \implies \mathbf{y = 1}$.
Collinear $\implies 2 - y = 2y - 1 \implies 3y = 3 \implies \mathbf{y = 1}$.
Q10. Prove by using slopes that $A(3, -1), B(-5, -5), C(1, 3)$ form a right-angled triangle:
Slope of $AB = \frac{-5 - (-1)}{-5-3} = \frac{-4}{-8} = \frac{1}{2}$.
Slope of $AC = \frac{3 - (-1)}{1-3} = \frac{4}{-2} = -2$.
Since $m_{AB} \cdot m_{AC} = \frac{1}{2} \cdot (-2) = -1$, $AB \perp AC$ at vertex $A$ ($\angle A = 90^\circ$). Thus $\Delta ABC$ is a right-angled triangle (Proved).
Slope of $AC = \frac{3 - (-1)}{1-3} = \frac{4}{-2} = -2$.
Since $m_{AB} \cdot m_{AC} = \frac{1}{2} \cdot (-2) = -1$, $AB \perp AC$ at vertex $A$ ($\angle A = 90^\circ$). Thus $\Delta ABC$ is a right-angled triangle (Proved).
Q11. Prove $A(-2, 1), B(6, 3), C(10, 5), D(2, 3)$ are vertices of a parallelogram:
Slope of $AB = \frac{3-1}{6 - (-2)} = \frac{2}{8} = \frac{1}{4}$; Slope of $CD = \frac{3-5}{2-10} = \frac{-2}{-8} = \frac{1}{4} \implies AB \parallel CD$.
Slope of $BC = \frac{5-3}{10-6} = \frac{2}{4} = \frac{1}{2}$; Slope of $DA = \frac{1-3}{-2-2} = \frac{-2}{-4} = \frac{1}{2} \implies BC \parallel DA$.
Opposite pairs are parallel $\implies ABCD$ is a parallelogram (Proved).
Slope of $BC = \frac{5-3}{10-6} = \frac{2}{4} = \frac{1}{2}$; Slope of $DA = \frac{1-3}{-2-2} = \frac{-2}{-4} = \frac{1}{2} \implies BC \parallel DA$.
Opposite pairs are parallel $\implies ABCD$ is a parallelogram (Proved).
Q12. $P(x, y), Q(-2, 2), R(1, 4), S(10, 1)$ are vertices of a parallelogram. Find $P(x, y)$:
In parallelogram $PQRS$, diagonals $PR$ and $QS$ bisect each other at midpoint $M$:
Midpoint of $QS = \left(\frac{-2 + 10}{2}, \frac{2 + 1}{2}\right) = \left(4, \frac{3}{2}\right)$.
Midpoint of $PR = \left(\frac{x + 1}{2}, \frac{y + 4}{2}\right) = \left(4, \frac{3}{2}\right) \implies \frac{x+1}{2} = 4 \implies x = 7$; $\frac{y+4}{2} = \frac{3}{2} \implies y = -1$.
Final Answer: $P(7, -1)$.
Midpoint of $QS = \left(\frac{-2 + 10}{2}, \frac{2 + 1}{2}\right) = \left(4, \frac{3}{2}\right)$.
Midpoint of $PR = \left(\frac{x + 1}{2}, \frac{y + 4}{2}\right) = \left(4, \frac{3}{2}\right) \implies \frac{x+1}{2} = 4 \implies x = 7$; $\frac{y+4}{2} = \frac{3}{2} \implies y = -1$.
Final Answer: $P(7, -1)$.
Q13. Three vertices of a rhombus are $A(2, -1), B(3, 4), C(-2, 3)$. Find fourth vertex $D(x, y)$:
Diagonals $AC$ and $BD$ bisect each other:
Midpoint of $AC = \left(\frac{2 - 2}{2}, \frac{-1 + 3}{2}\right) = (0, 1)$.
Midpoint of $BD = \left(\frac{3 + x}{2}, \frac{4 + y}{2}\right) = (0, 1) \implies 3 + x = 0 \implies x = -3$; $4 + y = 2 \implies y = -2$.
Final Answer: $D(-3, -2)$.
Midpoint of $AC = \left(\frac{2 - 2}{2}, \frac{-1 + 3}{2}\right) = (0, 1)$.
Midpoint of $BD = \left(\frac{3 + x}{2}, \frac{4 + y}{2}\right) = (0, 1) \implies 3 + x = 0 \implies x = -3$; $4 + y = 2 \implies y = -2$.
Final Answer: $D(-3, -2)$.
Q14. If $A(5, 0), B(0, 5), C(8, 8)$ are vertices of a triangle, find:
(i) Slopes of sides: $m_{AB} = \frac{5-0}{0-5} = \mathbf{-1}$; $m_{BC} = \frac{8-5}{8-0} = \mathbf{\frac{3}{8}}$; $m_{CA} = \frac{0-8}{5-8} = \frac{-8}{-3} = \mathbf{\frac{8}{3}}$.
(ii) Slopes of medians:
Midpoint of $BC = D\left(4, \frac{13}{2}\right) \implies \text{Slope}(AD) = \frac{13/2 - 0}{4 - 5} = \mathbf{-\frac{13}{2}}$.
Midpoint of $AC = E\left(\frac{13}{2}, 4\right) \implies \text{Slope}(BE) = \frac{4 - 5}{13/2 - 0} = \mathbf{-\frac{2}{13}}$.
Midpoint of $AB = F\left(\frac{5}{2}, \frac{5}{2}\right) \implies \text{Slope}(CF) = \frac{8 - 5/2}{8 - 5/2} = \frac{11/2}{11/2} = \mathbf{1}$.
(iii) Slopes of altitudes:
Altitude from $A \perp BC \implies m = -\frac{1}{3/8} = \mathbf{-\frac{8}{3}}$.
Altitude from $B \perp CA \implies m = -\frac{1}{8/3} = \mathbf{-\frac{3}{8}}$.
Altitude from $C \perp AB \implies m = -\frac{1}{-1} = \mathbf{1}$.
(ii) Slopes of medians:
Midpoint of $BC = D\left(4, \frac{13}{2}\right) \implies \text{Slope}(AD) = \frac{13/2 - 0}{4 - 5} = \mathbf{-\frac{13}{2}}$.
Midpoint of $AC = E\left(\frac{13}{2}, 4\right) \implies \text{Slope}(BE) = \frac{4 - 5}{13/2 - 0} = \mathbf{-\frac{2}{13}}$.
Midpoint of $AB = F\left(\frac{5}{2}, \frac{5}{2}\right) \implies \text{Slope}(CF) = \frac{8 - 5/2}{8 - 5/2} = \frac{11/2}{11/2} = \mathbf{1}$.
(iii) Slopes of altitudes:
Altitude from $A \perp BC \implies m = -\frac{1}{3/8} = \mathbf{-\frac{8}{3}}$.
Altitude from $B \perp CA \implies m = -\frac{1}{8/3} = \mathbf{-\frac{3}{8}}$.
Altitude from $C \perp AB \implies m = -\frac{1}{-1} = \mathbf{1}$.
Exercise 8.2 • Equations of Straight Lines (6 Standard Forms & Reductions)
Q1. Find equation of horizontal line ($y = b$) passing through:
(i) $(2, 3) \implies \mathbf{y = 3}$ |
(ii) $(8, 0) \implies \mathbf{y = 0}$ |
(iii) $(-5, -9) \implies \mathbf{y = -9}$ |
(iv) $\left(-\frac{3}{2}, -\frac{5}{2}\right) \implies \mathbf{y = -\frac{5}{2} \text{ (or } 2y + 5 = 0)}$
Q2. Find equation of vertical line ($x = a$) passing through:
(i) $(1, 5) \implies \mathbf{x = 1}$ |
(ii) $(9, 6) \implies \mathbf{x = 9}$ |
(iii) $(-4, -7) \implies \mathbf{x = -4}$ |
(iv) $\left(\frac{1}{4}, \frac{3}{4}\right) \implies \mathbf{x = \frac{1}{4} \text{ (or } 4x - 1 = 0)}$
Q3. Find equation of line with given conditions:
(i) Slope = 2, y-intercept = -3: $y = 2x - 3 \implies \mathbf{2x - y - 3 = 0}$
(ii) Through $(-5, 7)$ with slope 4: $y - 7 = 4(x + 5) \implies y - 7 = 4x + 20 \implies \mathbf{4x - y + 27 = 0}$
(iii) Through $(4, -5)$ with slope 0: $y - (-5) = 0(x - 4) \implies \mathbf{y + 5 = 0}$
(iv) Through $(-2, 9)$ with slope undefined: Vertical line $\implies \mathbf{x + 2 = 0}$
(v) Through $(-6, 1)$ and $(2, -4)$: $m = \frac{-4-1}{2 - (-6)} = -\frac{5}{8} \implies y - 1 = -\frac{5}{8}(x + 6) \implies 8y - 8 = -5x - 30 \implies \mathbf{5x + 8y + 22 = 0}$
(vi) Through $(2, -4)$ and $(8, 4)$: $m = \frac{4 - (-4)}{8-2} = \frac{8}{6} = \frac{4}{3} \implies y + 4 = \frac{4}{3}(x - 2) \implies 3y + 12 = 4x - 8 \implies \mathbf{4x - 3y - 20 = 0}$
(vii) $x$-intercept = -6, $y$-intercept = 5: $\frac{x}{-6} + \frac{y}{5} = 1 \implies -5x + 6y = 30 \implies \mathbf{5x - 6y + 30 = 0}$
(viii) Slope = -1, $x$-intercept = 11: Passes through $(11, 0) \implies y - 0 = -1(x - 11) \implies \mathbf{x + y - 11 = 0}$
(ii) Through $(-5, 7)$ with slope 4: $y - 7 = 4(x + 5) \implies y - 7 = 4x + 20 \implies \mathbf{4x - y + 27 = 0}$
(iii) Through $(4, -5)$ with slope 0: $y - (-5) = 0(x - 4) \implies \mathbf{y + 5 = 0}$
(iv) Through $(-2, 9)$ with slope undefined: Vertical line $\implies \mathbf{x + 2 = 0}$
(v) Through $(-6, 1)$ and $(2, -4)$: $m = \frac{-4-1}{2 - (-6)} = -\frac{5}{8} \implies y - 1 = -\frac{5}{8}(x + 6) \implies 8y - 8 = -5x - 30 \implies \mathbf{5x + 8y + 22 = 0}$
(vi) Through $(2, -4)$ and $(8, 4)$: $m = \frac{4 - (-4)}{8-2} = \frac{8}{6} = \frac{4}{3} \implies y + 4 = \frac{4}{3}(x - 2) \implies 3y + 12 = 4x - 8 \implies \mathbf{4x - 3y - 20 = 0}$
(vii) $x$-intercept = -6, $y$-intercept = 5: $\frac{x}{-6} + \frac{y}{5} = 1 \implies -5x + 6y = 30 \implies \mathbf{5x - 6y + 30 = 0}$
(viii) Slope = -1, $x$-intercept = 11: Passes through $(11, 0) \implies y - 0 = -1(x - 11) \implies \mathbf{x + y - 11 = 0}$
Q4. Symmetric Form $\frac{x - x_1}{\cos \theta} = \frac{y - y_1}{\sin \theta}$:
(i) $(-4, 2)$ and $\tan \theta = \frac{3}{4}$: $\sin \theta = \frac{3}{5}, \cos \theta = \frac{4}{5} \implies \mathbf{\frac{x + 4}{4/5} = \frac{y - 2}{3/5}}$ (or $\frac{x+4}{4} = \frac{y-2}{3}$)
(ii) $(6, -6)$ and $\theta = 30^\circ$: $\cos 30^\circ = \frac{\sqrt{3}}{2}, \sin 30^\circ = \frac{1}{2} \implies \mathbf{\frac{x - 6}{\sqrt{3}/2} = \frac{y + 6}{1/2}}$
(ii) $(6, -6)$ and $\theta = 30^\circ$: $\cos 30^\circ = \frac{\sqrt{3}}{2}, \sin 30^\circ = \frac{1}{2} \implies \mathbf{\frac{x - 6}{\sqrt{3}/2} = \frac{y + 6}{1/2}}$
Q5. Normal Form $x \cos \theta + y \sin \theta = p$:
(i) $p = 5, \theta = 120^\circ$: $\cos 120^\circ = -\frac{1}{2}, \sin 120^\circ = \frac{\sqrt{3}}{2} \implies \mathbf{-\frac{1}{2}x + \frac{\sqrt{3}}{2}y = 5 \text{ (or } -x + \sqrt{3}y = 10)}$
(ii) $p = 10, \tan \theta = 1 \implies \theta = 45^\circ$: $\cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}} \implies \mathbf{\frac{1}{\sqrt{2}}x + \frac{1}{\sqrt{2}}y = 10 \text{ (or } x + y = 10\sqrt{2})}$
(ii) $p = 10, \tan \theta = 1 \implies \theta = 45^\circ$: $\cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}} \implies \mathbf{\frac{1}{\sqrt{2}}x + \frac{1}{\sqrt{2}}y = 10 \text{ (or } x + y = 10\sqrt{2})}$
Q6. Find equation of straight line:
(i) Through $(-4, -4)$, parallel to slope $-5$: $y + 4 = -5(x + 4) \implies \mathbf{5x + y + 24 = 0}$
(ii) Through $(5, -1)$, perp to slope $\frac{1}{4}$: $m_{\perp} = -4 \implies y + 1 = -4(x - 5) \implies \mathbf{4x + y - 19 = 0}$
(iii) $y$-int = 4, parallel to slope $\frac{5}{2}$: $y = \frac{5}{2}x + 4 \implies \mathbf{5x - 2y + 8 = 0}$
(iv) $x$-int = -2, perp to slope 4: Passes $(-2, 0), m = -\frac{1}{4} \implies y - 0 = -\frac{1}{4}(x + 2) \implies \mathbf{x + 4y + 2 = 0}$
(v) Through $(-1, 4)$, perp to line through $(3, 0)$ and $(1, -2)$: $m_1 = \frac{-2-0}{1-3} = 1 \implies m_{\perp} = -1 \implies y - 4 = -1(x + 1) \implies \mathbf{x + y - 3 = 0}$
(vi) Through $(6, -4)$, parallel to line through $(-5, 2)$ and $(3, 6)$: $m = \frac{6-2}{3 - (-5)} = \frac{4}{8} = \frac{1}{2} \implies y + 4 = \frac{1}{2}(x - 6) \implies \mathbf{x - 2y - 14 = 0}$
(ii) Through $(5, -1)$, perp to slope $\frac{1}{4}$: $m_{\perp} = -4 \implies y + 1 = -4(x - 5) \implies \mathbf{4x + y - 19 = 0}$
(iii) $y$-int = 4, parallel to slope $\frac{5}{2}$: $y = \frac{5}{2}x + 4 \implies \mathbf{5x - 2y + 8 = 0}$
(iv) $x$-int = -2, perp to slope 4: Passes $(-2, 0), m = -\frac{1}{4} \implies y - 0 = -\frac{1}{4}(x + 2) \implies \mathbf{x + 4y + 2 = 0}$
(v) Through $(-1, 4)$, perp to line through $(3, 0)$ and $(1, -2)$: $m_1 = \frac{-2-0}{1-3} = 1 \implies m_{\perp} = -1 \implies y - 4 = -1(x + 1) \implies \mathbf{x + y - 3 = 0}$
(vi) Through $(6, -4)$, parallel to line through $(-5, 2)$ and $(3, 6)$: $m = \frac{6-2}{3 - (-5)} = \frac{4}{8} = \frac{1}{2} \implies y + 4 = \frac{1}{2}(x - 6) \implies \mathbf{x - 2y - 14 = 0}$
Q7. Line through $(3, 7)$ parallel to $4x - 3y + 1 = 0$:
$m = \frac{4}{3} \implies y - 7 = \frac{4}{3}(x - 3) \implies 3y - 21 = 4x - 12 \implies \mathbf{4x - 3y + 9 = 0}$.
Q8. Line through $(-2, -1)$ perp to $x - 2y = 0$:
Slope of $x - 2y = 0$ is $\frac{1}{2} \implies m_{\perp} = -2 \implies y + 1 = -2(x + 2) \implies \mathbf{2x + y + 5 = 0}$.
Q9. Perpendicular bisector of segment joining $(0, 6)$ and $(2, -2)$:
Midpoint $M = \left(\frac{0+2}{2}, \frac{6-2}{2}\right) = (1, 2)$.
Slope of segment $= \frac{-2-6}{2-0} = -4 \implies m_{\perp} = \frac{1}{4}$.
Equation: $y - 2 = \frac{1}{4}(x - 1) \implies 4y - 8 = x - 1 \implies \mathbf{x - 4y + 7 = 0}$.
Slope of segment $= \frac{-2-6}{2-0} = -4 \implies m_{\perp} = \frac{1}{4}$.
Equation: $y - 2 = \frac{1}{4}(x - 1) \implies 4y - 8 = x - 1 \implies \mathbf{x - 4y + 7 = 0}$.
Q10. Equations of medians and altitudes of $\Delta ABC$ with $A(0, 4), B(4, 6), C(-2, -2)$:
Medians:
- Median from $A$ through mid of $BC(1, 2)$: $m = \frac{2-4}{1-0} = -2 \implies \mathbf{2x + y - 4 = 0}$
- Median from $B$ through mid of $AC(-1, 1)$: $m = \frac{1-6}{-1-4} = 1 \implies y - 6 = 1(x - 4) \implies \mathbf{x - y + 2 = 0}$
- Median from $C$ through mid of $AB(2, 5)$: $m = \frac{5 - (-2)}{2 - (-2)} = \frac{7}{4} \implies \mathbf{7x - 4y + 6 = 0}$
Altitudes:
- Altitude from $A \perp BC$ ($m_{BC} = \frac{4}{3} \implies m_{\perp} = -\frac{3}{4}$): $y - 4 = -\frac{3}{4}(x - 0) \implies \mathbf{3x + 4y - 16 = 0}$
- Altitude from $B \perp AC$ ($m_{AC} = 3 \implies m_{\perp} = -\frac{1}{3}$): $y - 6 = -\frac{1}{3}(x - 4) \implies \mathbf{x + 3y - 22 = 0}$
- Altitude from $C \perp AB$ ($m_{AB} = \frac{1}{2} \implies m_{\perp} = -2$): $y + 2 = -2(x + 2) \implies \mathbf{2x + y + 6 = 0}$
- Median from $A$ through mid of $BC(1, 2)$: $m = \frac{2-4}{1-0} = -2 \implies \mathbf{2x + y - 4 = 0}$
- Median from $B$ through mid of $AC(-1, 1)$: $m = \frac{1-6}{-1-4} = 1 \implies y - 6 = 1(x - 4) \implies \mathbf{x - y + 2 = 0}$
- Median from $C$ through mid of $AB(2, 5)$: $m = \frac{5 - (-2)}{2 - (-2)} = \frac{7}{4} \implies \mathbf{7x - 4y + 6 = 0}$
Altitudes:
- Altitude from $A \perp BC$ ($m_{BC} = \frac{4}{3} \implies m_{\perp} = -\frac{3}{4}$): $y - 4 = -\frac{3}{4}(x - 0) \implies \mathbf{3x + 4y - 16 = 0}$
- Altitude from $B \perp AC$ ($m_{AC} = 3 \implies m_{\perp} = -\frac{1}{3}$): $y - 6 = -\frac{1}{3}(x - 4) \implies \mathbf{x + 3y - 22 = 0}$
- Altitude from $C \perp AB$ ($m_{AB} = \frac{1}{2} \implies m_{\perp} = -2$): $y + 2 = -2(x + 2) \implies \mathbf{2x + y + 6 = 0}$
Q11. Reduce equations into all standard forms:
(a) $6x + 8y - 11 = 0$:
- Slope-Intercept: $y = -\frac{3}{4}x + \frac{11}{8}$
- Two-Intercept: $\frac{x}{11/6} + \frac{y}{11/8} = 1$
- Point-Slope: $y - \frac{11}{8} = -\frac{3}{4}(x - 0)$
- Two-Point: $\frac{y - 11/8}{0 - 11/8} = \frac{x - 0}{11/6 - 0}$
- Normal Form: Divide by $\sqrt{6^2 + 8^2} = 10 \implies \mathbf{\frac{3}{5}x + \frac{4}{5}y = \frac{11}{10}}$ ($p = 1.1$)
- Symmetric Form: $\frac{x - 0}{4/5} = \frac{y - 11/8}{-3/5}$
(b) $4x - 3y + 9 = 0$:
- Slope-Intercept: $y = \frac{4}{3}x + 3$
- Two-Intercept: $\frac{x}{-9/4} + \frac{y}{3} = 1$
- Point-Slope: $y - 3 = \frac{4}{3}(x - 0)$
- Two-Point: $\frac{y - 3}{0 - 3} = \frac{x - 0}{-9/4 - 0}$
- Normal Form: $-4x + 3y = 9 \implies$ divide by $\sqrt{(-4)^2 + 3^2} = 5 \implies \mathbf{-\frac{4}{5}x + \frac{3}{5}y = \frac{9}{5}}$ ($p = 1.8$)
- Symmetric Form: $\frac{x - 0}{3/5} = \frac{y - 3}{4/5}$
- Slope-Intercept: $y = -\frac{3}{4}x + \frac{11}{8}$
- Two-Intercept: $\frac{x}{11/6} + \frac{y}{11/8} = 1$
- Point-Slope: $y - \frac{11}{8} = -\frac{3}{4}(x - 0)$
- Two-Point: $\frac{y - 11/8}{0 - 11/8} = \frac{x - 0}{11/6 - 0}$
- Normal Form: Divide by $\sqrt{6^2 + 8^2} = 10 \implies \mathbf{\frac{3}{5}x + \frac{4}{5}y = \frac{11}{10}}$ ($p = 1.1$)
- Symmetric Form: $\frac{x - 0}{4/5} = \frac{y - 11/8}{-3/5}$
(b) $4x - 3y + 9 = 0$:
- Slope-Intercept: $y = \frac{4}{3}x + 3$
- Two-Intercept: $\frac{x}{-9/4} + \frac{y}{3} = 1$
- Point-Slope: $y - 3 = \frac{4}{3}(x - 0)$
- Two-Point: $\frac{y - 3}{0 - 3} = \frac{x - 0}{-9/4 - 0}$
- Normal Form: $-4x + 3y = 9 \implies$ divide by $\sqrt{(-4)^2 + 3^2} = 5 \implies \mathbf{-\frac{4}{5}x + \frac{3}{5}y = \frac{9}{5}}$ ($p = 1.8$)
- Symmetric Form: $\frac{x - 0}{3/5} = \frac{y - 3}{4/5}$
Exercise 8.3 • Angles Between Lines, Intersection Points & Family of Lines
Q1. Measure of angle from $l_1$ to $l_2$ ($\tan \theta = \frac{m_2 - m_1}{1 + m_1 m_2}$):
(i) $m_1 = 0, m_2 = 1 \implies \tan \theta = \frac{1-0}{1+0} = 1 \implies \mathbf{\theta = 45^\circ}$
(ii) $m_1 = -0.5, m_2 = 4.5 \implies \tan \theta = \frac{4.5 - (-0.5)}{1 + (-0.5)(4.5)} = \frac{5}{1 - 2.25} = \frac{5}{-1.25} = -4 \implies \theta = 180^\circ - \tan^{-1}(4) \approx \mathbf{104.04^\circ}$ (or acute $75.96^\circ$)
(iii) $m_1 = \tan 45^\circ = 1, m_2 = \tan 135^\circ = -1 \implies m_1 m_2 = -1 \implies \mathbf{\theta = 90^\circ}$
(ii) $m_1 = -0.5, m_2 = 4.5 \implies \tan \theta = \frac{4.5 - (-0.5)}{1 + (-0.5)(4.5)} = \frac{5}{1 - 2.25} = \frac{5}{-1.25} = -4 \implies \theta = 180^\circ - \tan^{-1}(4) \approx \mathbf{104.04^\circ}$ (or acute $75.96^\circ$)
(iii) $m_1 = \tan 45^\circ = 1, m_2 = \tan 135^\circ = -1 \implies m_1 m_2 = -1 \implies \mathbf{\theta = 90^\circ}$
Q6. Find point of intersection:
(i) $2x + y + 1 = 0$ and $x - y - 4 = 0$: Adding gives $3x - 3 = 0 \implies x = 1 \implies y = -3$. Intersection: $\mathbf{(1, -3)}$
(ii) $x + y + 3 = 0$ and $2x - 5y + 8 = 0$: $x = -y - 3 \implies 2(-y-3) - 5y + 8 = 0 \implies -7y + 2 = 0 \implies y = \frac{2}{7}, x = -\frac{23}{7}$. Intersection: $\mathbf{\left(-\frac{23}{7}, \frac{2}{7}\right)}$
(iii) $2x + 5y + 3 = 0$ and $3x - 4y - 5 = 0$: Solving simultaneously gives $\mathbf{(1, -1)}$.
(ii) $x + y + 3 = 0$ and $2x - 5y + 8 = 0$: $x = -y - 3 \implies 2(-y-3) - 5y + 8 = 0 \implies -7y + 2 = 0 \implies y = \frac{2}{7}, x = -\frac{23}{7}$. Intersection: $\mathbf{\left(-\frac{23}{7}, \frac{2}{7}\right)}$
(iii) $2x + 5y + 3 = 0$ and $3x - 4y - 5 = 0$: Solving simultaneously gives $\mathbf{(1, -1)}$.
Q7. Family of lines through $(3x + 2y + 1) + k(x - 2y + 3) = 0$:
Intersection of $3x+2y+1=0$ and $x-2y+3=0$: adding gives $4x + 4 = 0 \implies x = -1, y = 1 \implies P(-1, 1)$.
(a) Through $(-1, 0)$: Slope $m = \frac{0-1}{-1 - (-1)} = \text{undefined} \implies$ Vertical line $\mathbf{x + 1 = 0}$.
(b) Parallel to $3x - 4y + 3 = 0$ ($m = \frac{3}{4}$): $y - 1 = \frac{3}{4}(x + 1) \implies \mathbf{3x - 4y + 7 = 0}$.
(a) Through $(-1, 0)$: Slope $m = \frac{0-1}{-1 - (-1)} = \text{undefined} \implies$ Vertical line $\mathbf{x + 1 = 0}$.
(b) Parallel to $3x - 4y + 3 = 0$ ($m = \frac{3}{4}$): $y - 1 = \frac{3}{4}(x + 1) \implies \mathbf{3x - 4y + 7 = 0}$.
Exercise 8.4 • Real-World Applied Coordinate Geometry Problems
Q1. Nasir fruiters & Shakri Malta:
Equation: $120x + 150y = 1200 \implies 4x + 5y = 40$.
If $y = 4$ dozen Malta $\implies 4x + 5(4) = 40 \implies 4x = 20 \implies \mathbf{x = 5 \text{ dozen fruiters}}$.
If $y = 4$ dozen Malta $\implies 4x + 5(4) = 40 \implies 4x = 20 \implies \mathbf{x = 5 \text{ dozen fruiters}}$.
Q2. Hotel total cost $y = 1450x + 2000$:
(i) 7 people: $y = 1450(7) + 2000 = 10150 + 2000 = \mathbf{\text{Rs. } 12,150}$.
(ii) Total cost Rs. 13,600: $13600 = 1450x + 2000 \implies 1450x = 11600 \implies \mathbf{x = 8 \text{ people}}$.
(ii) Total cost Rs. 13,600: $13600 = 1450x + 2000 \implies 1450x = 11600 \implies \mathbf{x = 8 \text{ people}}$.
Q7. Temperature scales ($^\circ C$ and $^\circ F$):
(i) $F = \frac{9}{5}C + 32$ (or $F = 1.8C + 32$).
(ii) $y$-intercept $= 32^\circ F$ (freezing point of water); slope $= \frac{9}{5} = 1.8$ (rate of change of $^\circ F$ per $^\circ C$).
(iii) At $5^\circ C$: $F = \frac{9}{5}(5) + 32 = 9 + 32 = \mathbf{41^\circ F}$.
(ii) $y$-intercept $= 32^\circ F$ (freezing point of water); slope $= \frac{9}{5} = 1.8$ (rate of change of $^\circ F$ per $^\circ C$).
(iii) At $5^\circ C$: $F = \frac{9}{5}(5) + 32 = 9 + 32 = \mathbf{41^\circ F}$.
Miscellaneous / Review Exercise 8 • Comprehensive Review & Objective Solutions
Q1. Textbook Multiple Choice Questions Key:
(i) Slopes 0 and $\infty \implies \mathbf{90^\circ}$ (c)
(ii) Right $\Delta$, angle $\implies \mathbf{45^\circ}$ (b)
(iii) Slope of $(-1, 6), (1, y)=3 \implies \mathbf{y = 12}$ (d)
(iv) $5x - ky - 3 = 0$ at $(1, 2) \implies \mathbf{k = 1}$ (a)
(v) $5x - 6 = 0 \implies \mathbf{\parallel \text{ to y-axis}}$ (b)
(vi) $y = 0 \implies \mathbf{\text{x-axis}}$ (a)
(vii) $y = b$ above x-axis $\implies \mathbf{b > 0}$ (c)
(viii) $x = a$ left to y-axis $\implies \mathbf{a < 0}$ (d)
(ix) Slope $-4 \implies \text{perp slope} = \mathbf{\frac{1}{4}}$ (a)
(x) Slope $\frac{2}{3} \implies \mathbf{2x - 3y = 2}$ (b)
(xi) $y = 5x - 3 \implies \mathbf{\text{slope-intercept}}$ (c)
(xii) $x + y = 5$, x-int $\implies \mathbf{5}$ (c)
(xiii) Intersects at $(2, 0), (0, 7) \implies \text{y-int } \mathbf{7}$ (d)
(xiv) Intersection of $9x-7y=0, 8x-11y=0 \implies \mathbf{(0, 0)}$ (a)
📝 Complete Solved Textbook Exercises & Examination Question Bank
Below is the exhaustive, step-by-step solution manual for every single textbook problem, example exercise, and review problem in Chapter 8, aligned strictly with FBISE scoring guidelines.
Exercise 8.1 • Step-by-Step Complete Solutions
Exercise 8.1 Q1 (i)
Exercise 8.1: Gradient from Inclination
: Find the gradient (slope) of the line whose inclination is $0^\circ$.
Detailed Step-by-Step Solution:
• Step 1: Formula: $m = \tan \theta$.
• Step 2: For $\theta = 0^\circ$, $m = \tan 0^\circ = 0$.
• Final Answer: m = 0
• Step 2: For $\theta = 0^\circ$, $m = \tan 0^\circ = 0$.
• Final Answer: m = 0
Exercise 8.1 Q1 (ii)
Exercise 8.1: Gradient from Inclination
: Find the gradient (slope) of the line whose inclination is $30^\circ$.
Detailed Step-by-Step Solution:
• Step 1: Formula: $m = \tan \theta$.
• Step 2: For $\theta = 30^\circ$, $m = \tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.577$.
• Final Answer: m = \frac{1}{\sqrt{3}} \approx 0.577
• Step 2: For $\theta = 30^\circ$, $m = \tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.577$.
• Final Answer: m = \frac{1}{\sqrt{3}} \approx 0.577
Exercise 8.1 Q1 (iii)
Exercise 8.1: Gradient from Inclination
: Find the gradient (slope) of the line whose inclination is $60^\circ$.
Detailed Step-by-Step Solution:
• Step 1: Formula: $m = \tan \theta$.
• Step 2: For $\theta = 60^\circ$, $m = \tan 60^\circ = \sqrt{3} \approx 1.732$.
• Final Answer: m = \sqrt{3} \approx 1.732
• Step 2: For $\theta = 60^\circ$, $m = \tan 60^\circ = \sqrt{3} \approx 1.732$.
• Final Answer: m = \sqrt{3} \approx 1.732
Exercise 8.1 Q1 (iv)
Exercise 8.1: Gradient from Inclination
: Find the gradient (slope) of the line whose inclination is $90^\circ$.
Detailed Step-by-Step Solution:
• Step 1: Formula: $m = \tan \theta$.
• Step 2: For $\theta = 90^\circ$, $m = \tan 90^\circ = \infty$ (Undefined, vertical line).
• Final Answer: m is Undefined
• Step 2: For $\theta = 90^\circ$, $m = \tan 90^\circ = \infty$ (Undefined, vertical line).
• Final Answer: m is Undefined
Exercise 8.1 Q1 (v)
Exercise 8.1: Gradient from Inclination
: Find the gradient (slope) of the line whose inclination is $120^\circ$.
Detailed Step-by-Step Solution:
• Step 1: $m = \tan 120^\circ = \tan(180^\circ - 60^\circ) = -\tan 60^\circ$.
• Step 2: $m = -\sqrt{3} \approx -1.732$.
• Final Answer: m = -\sqrt{3} \approx -1.732
• Step 2: $m = -\sqrt{3} \approx -1.732$.
• Final Answer: m = -\sqrt{3} \approx -1.732
Exercise 8.1 Q1 (vi)
Exercise 8.1: Gradient from Inclination
: Find the gradient (slope) of the line whose inclination is $150^\circ$.
Detailed Step-by-Step Solution:
• Step 1: $m = \tan 150^\circ = \tan(180^\circ - 30^\circ) = -\tan 30^\circ$.
• Step 2: $m = -\frac{1}{\sqrt{3}} \approx -0.577$.
• Final Answer: m = -\frac{1}{\sqrt{3}} \approx -0.577
• Step 2: $m = -\frac{1}{\sqrt{3}} \approx -0.577$.
• Final Answer: m = -\frac{1}{\sqrt{3}} \approx -0.577
Exercise 8.1 Q1 (vii)
Exercise 8.1: Gradient from Inclination
: Find the gradient (slope) of the line whose inclination is $170^\circ$.
Detailed Step-by-Step Solution:
• Step 1: $m = \tan 170^\circ = \tan(180^\circ - 10^\circ) = -\tan 10^\circ$.
• Step 2: $m \approx -0.1763$.
• Final Answer: m \approx -0.1763
• Step 2: $m \approx -0.1763$.
• Final Answer: m \approx -0.1763
Exercise 8.1 Q1 (viii)
Exercise 8.1: Gradient from Inclination
: Find the gradient (slope) of the line whose inclination is $45.5^\circ$.
Detailed Step-by-Step Solution:
• Step 1: $m = \tan(45.5^\circ)$.
• Step 2: Calculating value: $\tan(45.5^\circ) \approx 1.0176$.
• Final Answer: m \approx 1.0176
• Step 2: Calculating value: $\tan(45.5^\circ) \approx 1.0176$.
• Final Answer: m \approx 1.0176
Exercise 8.1 Q2 (i)
Exercise 8.1: Inclination from Slope
: Find the inclination of the line whose slope is $0$.
Detailed Step-by-Step Solution:
• Step 1: $m = \tan \theta = 0$.
• Step 2: $\theta = \tan^{-1}(0) = 0^\circ$.
• Final Answer: \theta = 0^\circ
• Step 2: $\theta = \tan^{-1}(0) = 0^\circ$.
• Final Answer: \theta = 0^\circ
Exercise 8.1 Q2 (ii)
Exercise 8.1: Inclination from Slope
: Find the inclination of the line whose slope is $0.577$.
Detailed Step-by-Step Solution:
• Step 1: $m = \tan \theta = 0.577 \approx \frac{1}{\sqrt{3}}$.
• Step 2: $\theta = \tan^{-1}(0.577) = 30^\circ$.
• Final Answer: \theta = 30^\circ
• Step 2: $\theta = \tan^{-1}(0.577) = 30^\circ$.
• Final Answer: \theta = 30^\circ
Exercise 8.1 Q2 (iii)
Exercise 8.1: Inclination from Slope
: Find the inclination of the line whose slope is $-1.732$.
Detailed Step-by-Step Solution:
• Step 1: $m = \tan \theta = -1.732 \approx -\sqrt{3}$.
• Step 2: $\theta = 180^\circ - \tan^{-1}(1.732) = 180^\circ - 60^\circ = 120^\circ$.
• Final Answer: \theta = 120^\circ
• Step 2: $\theta = 180^\circ - \tan^{-1}(1.732) = 180^\circ - 60^\circ = 120^\circ$.
• Final Answer: \theta = 120^\circ
Exercise 8.1 Q2 (iv)
Exercise 8.1: Inclination from Slope
: Find the inclination of the line whose slope is $-0.364$.
Detailed Step-by-Step Solution:
• Step 1: $m = \tan \theta = -0.364$.
• Step 2: Reference angle $\alpha = \tan^{-1}(0.364) = 20^\circ$.
• Step 3: Inclination $\theta = 180^\circ - 20^\circ = 160^\circ$.
• Final Answer: \theta = 160^\circ
• Step 2: Reference angle $\alpha = \tan^{-1}(0.364) = 20^\circ$.
• Step 3: Inclination $\theta = 180^\circ - 20^\circ = 160^\circ$.
• Final Answer: \theta = 160^\circ
Exercise 8.1 Q3 (i)
Exercise 8.1: Gradient & Inclination of Two Points
: Find the gradient and inclination of the line joining $A(2, 6)$ and $B(5, 8)$.
Detailed Step-by-Step Solution:
• Step 1: $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 6}{5 - 2} = \frac{2}{3}$.
• Step 2: $\theta = \tan^{-1}\left(\frac{2}{3}\right) = \tan^{-1}(0.6667) \approx 33.69^\circ$.
• Final Answer: m = 2/3, \theta \approx 33.69^\circ
• Step 2: $\theta = \tan^{-1}\left(\frac{2}{3}\right) = \tan^{-1}(0.6667) \approx 33.69^\circ$.
• Final Answer: m = 2/3, \theta \approx 33.69^\circ
Exercise 8.1 Q3 (ii)
Exercise 8.1: Gradient & Inclination of Two Points
: Find the gradient and inclination of the line joining $C(-2, 4)$ and $D(1, -3)$.
Detailed Step-by-Step Solution:
• Step 1: $m = \frac{-3 - 4}{1 - (-2)} = \frac{-7}{3} = -2.333$.
• Step 2: Reference angle $= \tan^{-1}(2.333) \approx 66.8^\circ$.
• Step 3: $\theta = 180^\circ - 66.8^\circ = 113.2^\circ$.
• Final Answer: m = -7/3, \theta \approx 113.2^\circ
• Step 2: Reference angle $= \tan^{-1}(2.333) \approx 66.8^\circ$.
• Step 3: $\theta = 180^\circ - 66.8^\circ = 113.2^\circ$.
• Final Answer: m = -7/3, \theta \approx 113.2^\circ
Exercise 8.1 Q3 (iii)
Exercise 8.1: Gradient & Inclination of Two Points
: Find the gradient and inclination of the line joining $E(5, -2)$ and $F(-2, -3)$.
Detailed Step-by-Step Solution:
• Step 1: $m = \frac{-3 - (-2)}{-2 - 5} = \frac{-1}{-7} = \frac{1}{7} \approx 0.1429$.
• Step 2: $\theta = \tan^{-1}(0.1429) \approx 8.13^\circ$.
• Final Answer: m = 1/7, \theta \approx 8.13^\circ
• Step 2: $\theta = \tan^{-1}(0.1429) \approx 8.13^\circ$.
• Final Answer: m = 1/7, \theta \approx 8.13^\circ
Exercise 8.1 Q4 (i)
Exercise 8.1: Parallel & Perpendicular Slopes
: If $A(-2, 6)$ and $B(7, -3)$, find the slope of the line parallel to $AB$.
Detailed Step-by-Step Solution:
• Step 1: $m_{AB} = \frac{-3 - 6}{7 - (-2)} = \frac{-9}{9} = -1$.
• Step 2: For parallel lines, $m_{\parallel} = m_{AB} = -1$.
• Final Answer: m = -1
• Step 2: For parallel lines, $m_{\parallel} = m_{AB} = -1$.
• Final Answer: m = -1
Exercise 8.1 Q4 (ii)
Exercise 8.1: Parallel & Perpendicular Slopes
: If $A(-2, 6)$ and $B(7, -3)$, find the slope of the line perpendicular to $AB$.
Detailed Step-by-Step Solution:
• Step 1: $m_{AB} = -1$.
• Step 2: For perpendicular lines, $m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{-1} = 1$.
• Final Answer: m = 1
• Step 2: For perpendicular lines, $m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{-1} = 1$.
• Final Answer: m = 1
Exercise 8.1 Q5
Exercise 8.1: Unknown Coordinate from Slope
: Find $x$ if the slope of the line passing through $A(3, x)$ and $B(5, 8)$ is $4$.
Detailed Step-by-Step Solution:
• Step 1: $m = \frac{8 - x}{5 - 3} = \frac{8 - x}{2}$.
• Step 2: Given $m = 4 \implies \frac{8 - x}{2} = 4 \implies 8 - x = 8 \implies x = 0$.
• Final Answer: x = 0
• Step 2: Given $m = 4 \implies \frac{8 - x}{2} = 4 \implies 8 - x = 8 \implies x = 0$.
• Final Answer: x = 0
Exercise 8.1 Q6
Exercise 8.1: Parallel Lines Condition
: Find $k$ if lines passing through $A(k, 2), B(3, 5)$ and $C(5, -1), D(8, 7)$ are parallel.
Detailed Step-by-Step Solution:
• Step 1: $m_1 = \frac{5 - 2}{3 - k} = \frac{3}{3 - k}$.
• Step 2: $m_2 = \frac{7 - (-1)}{8 - 5} = \frac{8}{3}$.
• Step 3: Parallel $\implies m_1 = m_2 \implies \frac{3}{3-k} = \frac{8}{3} \implies 9 = 24 - 8k \implies 8k = 15 \implies k = \frac{15}{8}$.
• Final Answer: k = 15/8
• Step 2: $m_2 = \frac{7 - (-1)}{8 - 5} = \frac{8}{3}$.
• Step 3: Parallel $\implies m_1 = m_2 \implies \frac{3}{3-k} = \frac{8}{3} \implies 9 = 24 - 8k \implies 8k = 15 \implies k = \frac{15}{8}$.
• Final Answer: k = 15/8
Exercise 8.1 Q7
Exercise 8.1: Perpendicular Lines Condition
: Find $k$ if lines passing through $P(-1, 2), Q(4, 7)$ and $R(2, k), S(7, 10)$ are perpendicular.
Detailed Step-by-Step Solution:
• Step 1: $m_1 = \frac{7 - 2}{4 - (-1)} = \frac{5}{5} = 1$.
• Step 2: $m_2 = \frac{10 - k}{7 - 2} = \frac{10 - k}{5}$.
• Step 3: Perpendicular $\implies m_1 \cdot m_2 = -1 \implies 1 \cdot \left(\frac{10 - k}{5}\right) = -1 \implies 10 - k = -5 \implies k = 15$.
• Final Answer: k = 15
• Step 2: $m_2 = \frac{10 - k}{7 - 2} = \frac{10 - k}{5}$.
• Step 3: Perpendicular $\implies m_1 \cdot m_2 = -1 \implies 1 \cdot \left(\frac{10 - k}{5}\right) = -1 \implies 10 - k = -5 \implies k = 15$.
• Final Answer: k = 15
Exercise 8.1 Q8
Exercise 8.1: Collinearity via Slopes
: Using slopes, prove that points $X(0, -3), Y(4, 7)$ and $Z(6, 12)$ are collinear.
Detailed Step-by-Step Solution:
• Step 1: Slope of $XY = \frac{7 - (-3)}{4 - 0} = \frac{10}{4} = \frac{5}{2}$.
• Step 2: Slope of $YZ = \frac{12 - 7}{6 - 4} = \frac{5}{2}$.
• Step 3: Since Slope($XY$) = Slope($YZ$) and $Y$ is a common point, points $X, Y, Z$ are collinear.
• Final Answer: Points X, Y, Z are Collinear (Proved)
• Step 2: Slope of $YZ = \frac{12 - 7}{6 - 4} = \frac{5}{2}$.
• Step 3: Since Slope($XY$) = Slope($YZ$) and $Y$ is a common point, points $X, Y, Z$ are collinear.
• Final Answer: Points X, Y, Z are Collinear (Proved)
Exercise 8.1 Q9
Exercise 8.1: Collinearity via Slopes
: Find the value of $y$ if points $P(4, y), Q(5, 2)$ and $R(6, 2y + 1)$ are collinear.
Detailed Step-by-Step Solution:
• Step 1: Slope of $PQ = \frac{2 - y}{5 - 4} = 2 - y$.
• Step 2: Slope of $QR = \frac{(2y + 1) - 2}{6 - 5} = 2y - 1$.
• Step 3: Collinear $\implies 2 - y = 2y - 1 \implies 3y = 3 \implies y = 1$.
• Final Answer: y = 1
• Step 2: Slope of $QR = \frac{(2y + 1) - 2}{6 - 5} = 2y - 1$.
• Step 3: Collinear $\implies 2 - y = 2y - 1 \implies 3y = 3 \implies y = 1$.
• Final Answer: y = 1
Exercise 8.1 Q10
Exercise 8.1: Right Triangle Proof via Slopes
: Prove by using slopes that points $A(3, -1), B(-5, -5)$ and $C(1, 3)$ are vertices of a right angled triangle.
Detailed Step-by-Step Solution:
• Step 1: $m_{AB} = \frac{-5 - (-1)}{-5 - 3} = \frac{-4}{-8} = \frac{1}{2}$.
• Step 2: $m_{AC} = \frac{3 - (-1)}{1 - 3} = \frac{4}{-2} = -2$.
• Step 3: Product of slopes: $m_{AB} \cdot m_{AC} = \frac{1}{2} \cdot (-2) = -1 \implies AB \perp AC$ at $\angle A = 90^\circ$.
• Final Answer: \Delta ABC is a Right-Angled Triangle (Proved)
• Step 2: $m_{AC} = \frac{3 - (-1)}{1 - 3} = \frac{4}{-2} = -2$.
• Step 3: Product of slopes: $m_{AB} \cdot m_{AC} = \frac{1}{2} \cdot (-2) = -1 \implies AB \perp AC$ at $\angle A = 90^\circ$.
• Final Answer: \Delta ABC is a Right-Angled Triangle (Proved)
Exercise 8.1 Q11
Exercise 8.1: Parallelogram Proof via Slopes
: Using slope, prove that $A(-2, 1), B(6, 3), C(10, 5)$ and $D(2, 3)$ are vertices of a parallelogram.
Detailed Step-by-Step Solution:
• Step 1: Slope of $AB = \frac{3 - 1}{6 - (-2)} = \frac{2}{8} = \frac{1}{4}$; Slope of $CD = \frac{3 - 5}{2 - 10} = \frac{-2}{-8} = \frac{1}{4} \implies AB \parallel CD$.
• Step 2: Slope of $BC = \frac{5 - 3}{10 - 6} = \frac{2}{4} = \frac{1}{2}$; Slope of $DA = \frac{1 - 3}{-2 - 2} = \frac{-2}{-4} = \frac{1}{2} \implies BC \parallel DA$.
• Step 3: Since both pairs of opposite sides are parallel, $ABCD$ is a parallelogram.
• Final Answer: ABCD is a Parallelogram (Proved)
• Step 2: Slope of $BC = \frac{5 - 3}{10 - 6} = \frac{2}{4} = \frac{1}{2}$; Slope of $DA = \frac{1 - 3}{-2 - 2} = \frac{-2}{-4} = \frac{1}{2} \implies BC \parallel DA$.
• Step 3: Since both pairs of opposite sides are parallel, $ABCD$ is a parallelogram.
• Final Answer: ABCD is a Parallelogram (Proved)
Exercise 8.1 Q12
Exercise 8.1: Parallelogram Missing Vertex
: $P(x, y), Q(-2, 2), R(1, 4)$ and $S(10, 1)$ are vertices of a parallelogram. Find $P(x, y)$.
Detailed Step-by-Step Solution:
• Step 1: In parallelogram $PQRS$, diagonals $PR$ and $QS$ bisect each other.
• Step 2: Midpoint of diagonal $QS = \left(\frac{-2 + 10}{2}, \frac{2 + 1}{2}\right) = \left(4, \frac{3}{2}\right)$.
• Step 3: Midpoint of diagonal $PR = \left(\frac{x + 1}{2}, \frac{y + 4}{2}\right) = \left(4, \frac{3}{2}\right)$.
• Step 4: $\frac{x + 1}{2} = 4 \implies x = 7$; $\frac{y + 4}{2} = \frac{3}{2} \implies y = -1$.
• Final Answer: P(7, -1)
• Step 2: Midpoint of diagonal $QS = \left(\frac{-2 + 10}{2}, \frac{2 + 1}{2}\right) = \left(4, \frac{3}{2}\right)$.
• Step 3: Midpoint of diagonal $PR = \left(\frac{x + 1}{2}, \frac{y + 4}{2}\right) = \left(4, \frac{3}{2}\right)$.
• Step 4: $\frac{x + 1}{2} = 4 \implies x = 7$; $\frac{y + 4}{2} = \frac{3}{2} \implies y = -1$.
• Final Answer: P(7, -1)
Exercise 8.1 Q13
Exercise 8.1: Rhombus Missing Vertex
: Three vertices of a rhombus are $A(2, -1), B(3, 4)$ and $C(-2, 3)$. Find the fourth vertex $D(x, y)$.
Detailed Step-by-Step Solution:
• Step 1: In rhombus $ABCD$, diagonals $AC$ and $BD$ bisect each other.
• Step 2: Midpoint of $AC = \left(\frac{2 - 2}{2}, \frac{-1 + 3}{2}\right) = (0, 1)$.
• Step 3: Midpoint of $BD = \left(\frac{3 + x}{2}, \frac{4 + y}{2}\right) = (0, 1)$.
• Step 4: $\frac{3 + x}{2} = 0 \implies x = -3$; $\frac{4 + y}{2} = 1 \implies y = -2$.
• Final Answer: D(-3, -2)
• Step 2: Midpoint of $AC = \left(\frac{2 - 2}{2}, \frac{-1 + 3}{2}\right) = (0, 1)$.
• Step 3: Midpoint of $BD = \left(\frac{3 + x}{2}, \frac{4 + y}{2}\right) = (0, 1)$.
• Step 4: $\frac{3 + x}{2} = 0 \implies x = -3$; $\frac{4 + y}{2} = 1 \implies y = -2$.
• Final Answer: D(-3, -2)
Exercise 8.1 Q14 (i)
Exercise 8.1: Triangle Slopes Analysis
: If $(5, 0), (0, 5)$ and $(8, 8)$ are vertices of a triangle, find the slopes of its sides.
Detailed Step-by-Step Solution:
• Step 1: Let $A(5, 0), B(0, 5), C(8, 8)$.
• Step 2: $m_{AB} = \frac{5 - 0}{0 - 5} = -1$.
• Step 3: $m_{BC} = \frac{8 - 5}{8 - 0} = \frac{3}{8}$.
• Step 4: $m_{CA} = \frac{0 - 8}{5 - 8} = \frac{-8}{-3} = \frac{8}{3}$.
• Final Answer: m_{AB} = -1, m_{BC} = 3/8, m_{CA} = 8/3
• Step 2: $m_{AB} = \frac{5 - 0}{0 - 5} = -1$.
• Step 3: $m_{BC} = \frac{8 - 5}{8 - 0} = \frac{3}{8}$.
• Step 4: $m_{CA} = \frac{0 - 8}{5 - 8} = \frac{-8}{-3} = \frac{8}{3}$.
• Final Answer: m_{AB} = -1, m_{BC} = 3/8, m_{CA} = 8/3
Exercise 8.1 Q14 (ii)
Exercise 8.1: Triangle Slopes Analysis
: If $A(5, 0), B(0, 5)$ and $C(8, 8)$ are vertices of a triangle, find the slopes of its medians.
Detailed Step-by-Step Solution:
• Step 1: Midpoint of $BC = D\left(4, \frac{13}{2}\right) \implies m_{AD} = \frac{13/2 - 0}{4 - 5} = -\frac{13}{2}$.
• Step 2: Midpoint of $AC = E\left(\frac{13}{2}, 4\right) \implies m_{BE} = \frac{4 - 5}{13/2 - 0} = -\frac{2}{13}$.
• Step 3: Midpoint of $AB = F\left(\frac{5}{2}, \frac{5}{2}\right) \implies m_{CF} = \frac{8 - 5/2}{8 - 5/2} = 1$.
• Final Answer: m_{AD} = -13/2, m_{BE} = -2/13, m_{CF} = 1
• Step 2: Midpoint of $AC = E\left(\frac{13}{2}, 4\right) \implies m_{BE} = \frac{4 - 5}{13/2 - 0} = -\frac{2}{13}$.
• Step 3: Midpoint of $AB = F\left(\frac{5}{2}, \frac{5}{2}\right) \implies m_{CF} = \frac{8 - 5/2}{8 - 5/2} = 1$.
• Final Answer: m_{AD} = -13/2, m_{BE} = -2/13, m_{CF} = 1
Exercise 8.1 Q14 (iii)
Exercise 8.1: Triangle Slopes Analysis
: If $A(5, 0), B(0, 5)$ and $C(8, 8)$ are vertices of a triangle, find the slopes of its altitudes.
Detailed Step-by-Step Solution:
• Step 1: Altitude from $A \perp BC \implies m_1 = -\frac{1}{m_{BC}} = -\frac{1}{3/8} = -\frac{8}{3}$.
• Step 2: Altitude from $B \perp CA \implies m_2 = -\frac{1}{m_{CA}} = -\frac{1}{8/3} = -\frac{3}{8}$.
• Step 3: Altitude from $C \perp AB \implies m_3 = -\frac{1}{m_{AB}} = -\frac{1}{-1} = 1$.
• Final Answer: m_1 = -8/3, m_2 = -3/8, m_3 = 1
• Step 2: Altitude from $B \perp CA \implies m_2 = -\frac{1}{m_{CA}} = -\frac{1}{8/3} = -\frac{3}{8}$.
• Step 3: Altitude from $C \perp AB \implies m_3 = -\frac{1}{m_{AB}} = -\frac{1}{-1} = 1$.
• Final Answer: m_1 = -8/3, m_2 = -3/8, m_3 = 1
Exercise 8.2 • Step-by-Step Complete Solutions
Exercise 8.2 Q1 (i)
Exercise 8.2: Horizontal Line Equations
: Find the equation of the horizontal line passing through $(2, 3)$.
Detailed Step-by-Step Solution:
• Step 1: A horizontal line has slope $m = 0$ and equation $y = b$.
• Step 2: Through $(2, 3)$, $y = 3$.
• Final Answer: y = 3
• Step 2: Through $(2, 3)$, $y = 3$.
• Final Answer: y = 3
Exercise 8.2 Q1 (ii)
Exercise 8.2: Horizontal Line Equations
: Find the equation of the horizontal line passing through $(8, 0)$.
Detailed Step-by-Step Solution:
• Step 1: Horizontal line equation is $y = b$.
• Step 2: Through $(8, 0)$, $y = 0$ (the x-axis).
• Final Answer: y = 0
• Step 2: Through $(8, 0)$, $y = 0$ (the x-axis).
• Final Answer: y = 0
Exercise 8.2 Q1 (iii)
Exercise 8.2: Horizontal Line Equations
: Find the equation of the horizontal line passing through $(-5, -9)$.
Detailed Step-by-Step Solution:
• Step 1: Horizontal line equation is $y = b$.
• Step 2: Through $(-5, -9)$, $y = -9$ (or $y + 9 = 0$).
• Final Answer: y = -9
• Step 2: Through $(-5, -9)$, $y = -9$ (or $y + 9 = 0$).
• Final Answer: y = -9
Exercise 8.2 Q1 (iv)
Exercise 8.2: Horizontal Line Equations
: Find the equation of the horizontal line passing through $\left(-\frac{3}{2}, -\frac{5}{2}\right)$.
Detailed Step-by-Step Solution:
• Step 1: Horizontal line equation is $y = b$.
• Step 2: $y = -\frac{5}{2} \implies 2y + 5 = 0$.
• Final Answer: 2y + 5 = 0
• Step 2: $y = -\frac{5}{2} \implies 2y + 5 = 0$.
• Final Answer: 2y + 5 = 0
Exercise 8.2 Q2 (i)
Exercise 8.2: Vertical Line Equations
: Find the equation of the vertical line passing through $(1, 5)$.
Detailed Step-by-Step Solution:
• Step 1: A vertical line has undefined slope and equation $x = a$.
• Step 2: Through $(1, 5)$, $x = 1$.
• Final Answer: x = 1
• Step 2: Through $(1, 5)$, $x = 1$.
• Final Answer: x = 1
Exercise 8.2 Q2 (ii)
Exercise 8.2: Vertical Line Equations
: Find the equation of the vertical line passing through $(9, 6)$.
Detailed Step-by-Step Solution:
• Step 1: Vertical line equation is $x = a$.
• Step 2: Through $(9, 6)$, $x = 9$.
• Final Answer: x = 9
• Step 2: Through $(9, 6)$, $x = 9$.
• Final Answer: x = 9
Exercise 8.2 Q2 (iii)
Exercise 8.2: Vertical Line Equations
: Find the equation of the vertical line passing through $(-4, -7)$.
Detailed Step-by-Step Solution:
• Step 1: Vertical line equation is $x = a$.
• Step 2: Through $(-4, -7)$, $x = -4$ (or $x + 4 = 0$).
• Final Answer: x = -4
• Step 2: Through $(-4, -7)$, $x = -4$ (or $x + 4 = 0$).
• Final Answer: x = -4
Exercise 8.2 Q2 (iv)
Exercise 8.2: Vertical Line Equations
: Find the equation of the vertical line passing through $\left(\frac{1}{4}, \frac{3}{4}\right)$.
Detailed Step-by-Step Solution:
• Step 1: Vertical line equation is $x = a$.
• Step 2: $x = \frac{1}{4} \implies 4x - 1 = 0$.
• Final Answer: 4x - 1 = 0
• Step 2: $x = \frac{1}{4} \implies 4x - 1 = 0$.
• Final Answer: 4x - 1 = 0
Exercise 8.2 Q3 (i)
Exercise 8.2: Standard Line Formations
: Find the equation of the line with $\text{slope} = 2$ and $y\text{-intercept} = -3$.
Detailed Step-by-Step Solution:
• Step 1: Slope-intercept form: $y = mx + c$.
• Step 2: $y = 2x + (-3) \implies 2x - y - 3 = 0$.
• Final Answer: 2x - y - 3 = 0
• Step 2: $y = 2x + (-3) \implies 2x - y - 3 = 0$.
• Final Answer: 2x - y - 3 = 0
Exercise 8.2 Q3 (ii)
Exercise 8.2: Standard Line Formations
: Find the equation of the line passing through $(-5, 7)$ with slope $4$.
Detailed Step-by-Step Solution:
• Step 1: Point-slope form: $y - y_1 = m(x - x_1)$.
• Step 2: $y - 7 = 4(x - (-5)) \implies y - 7 = 4x + 20 \implies 4x - y + 27 = 0$.
• Final Answer: 4x - y + 27 = 0
• Step 2: $y - 7 = 4(x - (-5)) \implies y - 7 = 4x + 20 \implies 4x - y + 27 = 0$.
• Final Answer: 4x - y + 27 = 0
Exercise 8.2 Q3 (iii)
Exercise 8.2: Standard Line Formations
: Find the equation of the line passing through $(4, -5)$ with slope $0$.
Detailed Step-by-Step Solution:
• Step 1: $y - (-5) = 0(x - 4) \implies y + 5 = 0$.
• Final Answer: y + 5 = 0
• Final Answer: y + 5 = 0
Exercise 8.2 Q3 (iv)
Exercise 8.2: Standard Line Formations
: Find the equation of the line passing through $(-2, 9)$ with slope undefined.
Detailed Step-by-Step Solution:
• Step 1: Undefined slope implies a vertical line $x = a$.
• Step 2: $x = -2 \implies x + 2 = 0$.
• Final Answer: x + 2 = 0
• Step 2: $x = -2 \implies x + 2 = 0$.
• Final Answer: x + 2 = 0
Exercise 8.2 Q3 (v)
Exercise 8.2: Standard Line Formations
: Find the equation of the line passing through $(-6, 1)$ and $(2, -4)$.
Detailed Step-by-Step Solution:
• Step 1: $m = \frac{-4 - 1}{2 - (-6)} = -\frac{5}{8}$.
• Step 2: $y - 1 = -\frac{5}{8}(x + 6) \implies 8y - 8 = -5x - 30 \implies 5x + 8y + 22 = 0$.
• Final Answer: 5x + 8y + 22 = 0
• Step 2: $y - 1 = -\frac{5}{8}(x + 6) \implies 8y - 8 = -5x - 30 \implies 5x + 8y + 22 = 0$.
• Final Answer: 5x + 8y + 22 = 0
Exercise 8.2 Q3 (vi)
Exercise 8.2: Standard Line Formations
: Find the equation of the line passing through $(2, -4)$ and $(8, 4)$.
Detailed Step-by-Step Solution:
• Step 1: $m = \frac{4 - (-4)}{8 - 2} = \frac{8}{6} = \frac{4}{3}$.
• Step 2: $y - 4 = \frac{4}{3}(x - 8) \implies 3y - 12 = 4x - 32 \implies 4x - 3y - 20 = 0$.
• Final Answer: 4x - 3y - 20 = 0
• Step 2: $y - 4 = \frac{4}{3}(x - 8) \implies 3y - 12 = 4x - 32 \implies 4x - 3y - 20 = 0$.
• Final Answer: 4x - 3y - 20 = 0
Exercise 8.2 Q3 (vii)
Exercise 8.2: Standard Line Formations
: Find the equation of the line with $x\text{-intercept} = -6$ and $y\text{-intercept} = 5$.
Detailed Step-by-Step Solution:
• Step 1: Two-intercept form: $\frac{x}{a} + \frac{y}{b} = 1$.
• Step 2: $\frac{x}{-6} + \frac{y}{5} = 1 \implies -5x + 6y = 30 \implies 5x - 6y + 30 = 0$.
• Final Answer: 5x - 6y + 30 = 0
• Step 2: $\frac{x}{-6} + \frac{y}{5} = 1 \implies -5x + 6y = 30 \implies 5x - 6y + 30 = 0$.
• Final Answer: 5x - 6y + 30 = 0
Exercise 8.2 Q3 (viii)
Exercise 8.2: Standard Line Formations
: Find the equation of the line with $\text{slope} = -1$ and $x\text{-intercept} = 11$.
Detailed Step-by-Step Solution:
• Step 1: $x$-intercept is $11 \implies$ point is $(11, 0)$.
• Step 2: $y - 0 = -1(x - 11) \implies y = -x + 11 \implies x + y - 11 = 0$.
• Final Answer: x + y - 11 = 0
• Step 2: $y - 0 = -1(x - 11) \implies y = -x + 11 \implies x + y - 11 = 0$.
• Final Answer: x + y - 11 = 0
Exercise 8.2 Q4 (i)
Exercise 8.2: Symmetric Form of Line
: Find equation of line in symmetric form when $(x_1, y_1) = (-4, 2)$ and $\tan \theta = \frac{3}{4}$.
Detailed Step-by-Step Solution:
• Step 1: $\tan \theta = \frac{3}{4} \implies \sin \theta = \frac{3}{5}, \cos \theta = \frac{4}{5}$.
• Step 2: Symmetric form: $\frac{x - x_1}{\cos \theta} = \frac{y - y_1}{\sin \theta}$.
• Step 3: $\frac{x + 4}{4/5} = \frac{y - 2}{3/5}$ (or $\frac{x+4}{4} = \frac{y-2}{3}$).
• Final Answer: \frac{x + 4}{4/5} = \frac{y - 2}{3/5}
• Step 2: Symmetric form: $\frac{x - x_1}{\cos \theta} = \frac{y - y_1}{\sin \theta}$.
• Step 3: $\frac{x + 4}{4/5} = \frac{y - 2}{3/5}$ (or $\frac{x+4}{4} = \frac{y-2}{3}$).
• Final Answer: \frac{x + 4}{4/5} = \frac{y - 2}{3/5}
Exercise 8.2 Q4 (ii)
Exercise 8.2: Symmetric Form of Line
: Find equation of line in symmetric form when $(x_1, y_1) = (6, -6)$ and $\theta = 30^\circ$.
Detailed Step-by-Step Solution:
• Step 1: $\cos 30^\circ = \frac{\sqrt{3}}{2}, \sin 30^\circ = \frac{1}{2}$.
• Step 2: Symmetric form: $\frac{x - 6}{\sqrt{3}/2} = \frac{y - (-6)}{1/2} = \frac{y + 6}{1/2}$.
• Final Answer: \frac{x - 6}{\sqrt{3}/2} = \frac{y + 6}{1/2}
• Step 2: Symmetric form: $\frac{x - 6}{\sqrt{3}/2} = \frac{y - (-6)}{1/2} = \frac{y + 6}{1/2}$.
• Final Answer: \frac{x - 6}{\sqrt{3}/2} = \frac{y + 6}{1/2}
Exercise 8.2 Q5 (i)
Exercise 8.2: Normal Form of Line
: Find equation of line in normal form when $p = 5$ and $\theta = 120^\circ$.
Detailed Step-by-Step Solution:
• Step 1: Normal form: $x \cos \theta + y \sin \theta = p$.
• Step 2: $\cos 120^\circ = -\frac{1}{2}, \sin 120^\circ = \frac{\sqrt{3}}{2}$.
• Step 3: $-\frac{1}{2}x + \frac{\sqrt{3}}{2}y = 5$ (or $-x + \sqrt{3}y = 10$).
• Final Answer: -\frac{1}{2}x + \frac{\sqrt{3}}{2}y = 5
• Step 2: $\cos 120^\circ = -\frac{1}{2}, \sin 120^\circ = \frac{\sqrt{3}}{2}$.
• Step 3: $-\frac{1}{2}x + \frac{\sqrt{3}}{2}y = 5$ (or $-x + \sqrt{3}y = 10$).
• Final Answer: -\frac{1}{2}x + \frac{\sqrt{3}}{2}y = 5
Exercise 8.2 Q5 (ii)
Exercise 8.2: Normal Form of Line
: Find equation of line in normal form when $p = 10$ and $\tan \theta = 1$.
Detailed Step-by-Step Solution:
• Step 1: $\tan \theta = 1 \implies \theta = 45^\circ \implies \cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}}$.
• Step 2: Normal form: $\frac{1}{\sqrt{2}}x + \frac{1}{\sqrt{2}}y = 10$ (or $x + y = 10\sqrt{2}$).
• Final Answer: \frac{1}{\sqrt{2}}x + \frac{1}{\sqrt{2}}y = 10
• Step 2: Normal form: $\frac{1}{\sqrt{2}}x + \frac{1}{\sqrt{2}}y = 10$ (or $x + y = 10\sqrt{2}$).
• Final Answer: \frac{1}{\sqrt{2}}x + \frac{1}{\sqrt{2}}y = 10
Exercise 8.2 Q6 (i)
Exercise 8.2: Line Equations with Geometric Conditions
: Find the equation of straight line through $(-4, -4)$ and parallel to the line with slope $-5$.
Detailed Step-by-Step Solution:
• Step 1: Parallel slope $m = -5$.
• Step 2: $y - (-4) = -5(x - (-4)) \implies y + 4 = -5x - 20 \implies 5x + y + 24 = 0$.
• Final Answer: 5x + y + 24 = 0
• Step 2: $y - (-4) = -5(x - (-4)) \implies y + 4 = -5x - 20 \implies 5x + y + 24 = 0$.
• Final Answer: 5x + y + 24 = 0
Exercise 8.2 Q6 (ii)
Exercise 8.2: Line Equations with Geometric Conditions
: Find the equation of straight line through $(5, -1)$ and perpendicular to the line with slope $\frac{1}{4}$.
Detailed Step-by-Step Solution:
• Step 1: Perpendicular slope $m = -\frac{1}{1/4} = -4$.
• Step 2: $y - (-1) = -4(x - 5) \implies y + 1 = -4x + 20 \implies 4x + y - 19 = 0$.
• Final Answer: 4x + y - 19 = 0
• Step 2: $y - (-1) = -4(x - 5) \implies y + 1 = -4x + 20 \implies 4x + y - 19 = 0$.
• Final Answer: 4x + y - 19 = 0
Exercise 8.2 Q6 (iii)
Exercise 8.2: Line Equations with Geometric Conditions
: Find the equation of straight line having $y\text{-intercept} = 4$ and parallel to the line with slope $\frac{5}{2}$.
Detailed Step-by-Step Solution:
• Step 1: $m = \frac{5}{2}, c = 4$.
• Step 2: $y = \frac{5}{2}x + 4 \implies 2y = 5x + 8 \implies 5x - 2y + 8 = 0$.
• Final Answer: 5x - 2y + 8 = 0
• Step 2: $y = \frac{5}{2}x + 4 \implies 2y = 5x + 8 \implies 5x - 2y + 8 = 0$.
• Final Answer: 5x - 2y + 8 = 0
Exercise 8.2 Q6 (iv)
Exercise 8.2: Line Equations with Geometric Conditions
: Find the equation of straight line having $x\text{-intercept} = -2$ and perpendicular to the line with slope $4$.
Detailed Step-by-Step Solution:
• Step 1: $x$-intercept is $-2 \implies (-2, 0)$. Perpendicular slope $m = -\frac{1}{4}$.
• Step 2: $y - 0 = -\frac{1}{4}(x + 2) \implies 4y = -x - 2 \implies x + 4y + 2 = 0$.
• Final Answer: x + 4y + 2 = 0
• Step 2: $y - 0 = -\frac{1}{4}(x + 2) \implies 4y = -x - 2 \implies x + 4y + 2 = 0$.
• Final Answer: x + 4y + 2 = 0
Exercise 8.2 Q6 (v)
Exercise 8.2: Line Equations with Geometric Conditions
: Find the equation of straight line through $(-1, 4)$ and perpendicular to the line passing through $(3, 0)$ and $(1, -2)$.
Detailed Step-by-Step Solution:
• Step 1: Slope of given line $= \frac{-2 - 0}{1 - 3} = \frac{-2}{-2} = 1$.
• Step 2: Perpendicular slope $m_{\perp} = -1$.
• Step 3: $y - 4 = -1(x - (-1)) \implies y - 4 = -x - 1 \implies x + y - 3 = 0$.
• Final Answer: x + y - 3 = 0
• Step 2: Perpendicular slope $m_{\perp} = -1$.
• Step 3: $y - 4 = -1(x - (-1)) \implies y - 4 = -x - 1 \implies x + y - 3 = 0$.
• Final Answer: x + y - 3 = 0
Exercise 8.2 Q6 (vi)
Exercise 8.2: Line Equations with Geometric Conditions
: Find the equation of straight line through $(6, -4)$ and parallel to the line passing through $(-5, 2)$ and $(3, 6)$.
Detailed Step-by-Step Solution:
• Step 1: Slope of given line $= \frac{6 - 2}{3 - (-5)} = \frac{4}{8} = \frac{1}{2}$.
• Step 2: Parallel slope $m = \frac{1}{2}$.
• Step 3: $y - (-4) = \frac{1}{2}(x - 6) \implies 2y + 8 = x - 6 \implies x - 2y - 14 = 0$.
• Final Answer: x - 2y - 14 = 0
• Step 2: Parallel slope $m = \frac{1}{2}$.
• Step 3: $y - (-4) = \frac{1}{2}(x - 6) \implies 2y + 8 = x - 6 \implies x - 2y - 14 = 0$.
• Final Answer: x - 2y - 14 = 0
Exercise 8.2 Q7
Exercise 8.2: Parallel Line Equation
: Find the equation of the line through $(3, 7)$ and parallel to the line $4x - 3y + 1 = 0$.
Detailed Step-by-Step Solution:
• Step 1: Slope of $4x - 3y + 1 = 0$ is $m = \frac{4}{3}$.
• Step 2: Parallel line through $(3, 7)$: $y - 7 = \frac{4}{3}(x - 3) \implies 3y - 21 = 4x - 12 \implies 4x - 3y + 9 = 0$.
• Final Answer: 4x - 3y + 9 = 0
• Step 2: Parallel line through $(3, 7)$: $y - 7 = \frac{4}{3}(x - 3) \implies 3y - 21 = 4x - 12 \implies 4x - 3y + 9 = 0$.
• Final Answer: 4x - 3y + 9 = 0
Exercise 8.2 Q8
Exercise 8.2: Perpendicular Line Equation
: Find the equation of the line through $(-2, -1)$ and perpendicular to the line $x - 2y = 0$.
Detailed Step-by-Step Solution:
• Step 1: Slope of $x - 2y = 0$ is $\frac{1}{2} \implies m_{\perp} = -2$.
• Step 2: $y - (-1) = -2(x - (-2)) \implies y + 1 = -2x - 4 \implies 2x + y + 5 = 0$.
• Final Answer: 2x + y + 5 = 0
• Step 2: $y - (-1) = -2(x - (-2)) \implies y + 1 = -2x - 4 \implies 2x + y + 5 = 0$.
• Final Answer: 2x + y + 5 = 0
Exercise 8.2 Q9
Exercise 8.2: Perpendicular Bisector
: Find the equation of the perpendicular bisector of the line segment joining $(0, 6)$ and $(2, -2)$.
Detailed Step-by-Step Solution:
• Step 1: Midpoint $M = \left(\frac{0 + 2}{2}, \frac{6 - 2}{2}\right) = (1, 2)$.
• Step 2: Slope of segment $= \frac{-2 - 6}{2 - 0} = -4 \implies m_{\perp} = \frac{1}{4}$.
• Step 3: $y - 2 = \frac{1}{4}(x - 1) \implies 4y - 8 = x - 1 \implies x - 4y + 7 = 0$.
• Final Answer: x - 4y + 7 = 0
• Step 2: Slope of segment $= \frac{-2 - 6}{2 - 0} = -4 \implies m_{\perp} = \frac{1}{4}$.
• Step 3: $y - 2 = \frac{1}{4}(x - 1) \implies 4y - 8 = x - 1 \implies x - 4y + 7 = 0$.
• Final Answer: x - 4y + 7 = 0
Exercise 8.2 Q10
Exercise 8.2: Medians & Altitudes Equations
: Find equations of medians and altitudes of triangle with vertices $A(0, 4), B(4, 6)$ and $C(-2, -2)$.
Detailed Step-by-Step Solution:
• Step 1 (Medians): Midpoints are $D(1, 2), E(-1, 1), F(2, 5)$.
- Median $AD$: $m = \frac{2-4}{1-0} = -2 \implies 2x + y - 4 = 0$.
- Median $BE$: $m = \frac{1-6}{-1-4} = 1 \implies x - y + 2 = 0$.
- Median $CF$: $m = \frac{5-(-2)}{2-(-2)} = \frac{7}{4} \implies 7x - 4y + 6 = 0$.
• Step 2 (Altitudes):
- $m_{BC} = \frac{4}{3} \implies m_{\perp} = -\frac{3}{4} \implies y - 4 = -\frac{3}{4}x \implies 3x + 4y - 16 = 0$.
- $m_{AC} = 3 \implies m_{\perp} = -\frac{1}{3} \implies y - 6 = -\frac{1}{3}(x - 4) \implies x + 3y - 22 = 0$.
- $m_{AB} = \frac{1}{2} \implies m_{\perp} = -2 \implies y + 2 = -2(x + 2) \implies 2x + y + 6 = 0$.
• Final Answer: Medians: 2x+y-4=0, x-y+2=0, 7x-4y+6=0; Altitudes: 3x+4y-16=0, x+3y-22=0, 2x+y+6=0
- Median $AD$: $m = \frac{2-4}{1-0} = -2 \implies 2x + y - 4 = 0$.
- Median $BE$: $m = \frac{1-6}{-1-4} = 1 \implies x - y + 2 = 0$.
- Median $CF$: $m = \frac{5-(-2)}{2-(-2)} = \frac{7}{4} \implies 7x - 4y + 6 = 0$.
• Step 2 (Altitudes):
- $m_{BC} = \frac{4}{3} \implies m_{\perp} = -\frac{3}{4} \implies y - 4 = -\frac{3}{4}x \implies 3x + 4y - 16 = 0$.
- $m_{AC} = 3 \implies m_{\perp} = -\frac{1}{3} \implies y - 6 = -\frac{1}{3}(x - 4) \implies x + 3y - 22 = 0$.
- $m_{AB} = \frac{1}{2} \implies m_{\perp} = -2 \implies y + 2 = -2(x + 2) \implies 2x + y + 6 = 0$.
• Final Answer: Medians: 2x+y-4=0, x-y+2=0, 7x-4y+6=0; Altitudes: 3x+4y-16=0, x+3y-22=0, 2x+y+6=0
Exercise 8.2 Q11 (a)
Exercise 8.2: Reduction of General Form
: Reduce the equation $6x + 8y - 11 = 0$ into all 6 standard forms.
Detailed Step-by-Step Solution:
• Step 1 (Slope-Intercept): $8y = -6x + 11 \implies y = -\frac{3}{4}x + \frac{11}{8}$.
• Step 2 (Two-Intercept): $6x + 8y = 11 \implies \frac{x}{11/6} + \frac{y}{11/8} = 1$.
• Step 3 (Point-Slope): $y - \frac{11}{8} = -\frac{3}{4}(x - 0)$.
• Step 4 (Two-Point): $\frac{y - 11/8}{0 - 11/8} = \frac{x - 0}{11/6 - 0}$.
• Step 5 (Normal Form): Divide by $\sqrt{6^2 + 8^2} = 10 \implies \frac{3}{5}x + \frac{4}{5}y = \frac{11}{10}$.
• Step 6 (Symmetric Form): $\frac{x - 0}{4/5} = \frac{y - 11/8}{-3/5}$.
• Final Answer: Reduced into 6 standard forms
• Step 2 (Two-Intercept): $6x + 8y = 11 \implies \frac{x}{11/6} + \frac{y}{11/8} = 1$.
• Step 3 (Point-Slope): $y - \frac{11}{8} = -\frac{3}{4}(x - 0)$.
• Step 4 (Two-Point): $\frac{y - 11/8}{0 - 11/8} = \frac{x - 0}{11/6 - 0}$.
• Step 5 (Normal Form): Divide by $\sqrt{6^2 + 8^2} = 10 \implies \frac{3}{5}x + \frac{4}{5}y = \frac{11}{10}$.
• Step 6 (Symmetric Form): $\frac{x - 0}{4/5} = \frac{y - 11/8}{-3/5}$.
• Final Answer: Reduced into 6 standard forms
Exercise 8.2 Q11 (b)
Exercise 8.2: Reduction of General Form
: Reduce the equation $4x - 3y + 9 = 0$ into all 6 standard forms.
Detailed Step-by-Step Solution:
• Step 1 (Slope-Intercept): $3y = 4x + 9 \implies y = \frac{4}{3}x + 3$.
• Step 2 (Two-Intercept): $4x - 3y = -9 \implies \frac{x}{-9/4} + \frac{y}{3} = 1$.
• Step 3 (Point-Slope): $y - 3 = \frac{4}{3}(x - 0)$.
• Step 4 (Two-Point): $\frac{y - 3}{0 - 3} = \frac{x - 0}{-9/4 - 0}$.
• Step 5 (Normal Form): $-4x + 3y = 9 \implies$ divide by $\sqrt{(-4)^2 + 3^2} = 5 \implies -\frac{4}{5}x + \frac{3}{5}y = \frac{9}{5}$.
• Step 6 (Symmetric Form): $\frac{x - 0}{3/5} = \frac{y - 3}{4/5}$.
• Final Answer: Reduced into 6 standard forms
• Step 2 (Two-Intercept): $4x - 3y = -9 \implies \frac{x}{-9/4} + \frac{y}{3} = 1$.
• Step 3 (Point-Slope): $y - 3 = \frac{4}{3}(x - 0)$.
• Step 4 (Two-Point): $\frac{y - 3}{0 - 3} = \frac{x - 0}{-9/4 - 0}$.
• Step 5 (Normal Form): $-4x + 3y = 9 \implies$ divide by $\sqrt{(-4)^2 + 3^2} = 5 \implies -\frac{4}{5}x + \frac{3}{5}y = \frac{9}{5}$.
• Step 6 (Symmetric Form): $\frac{x - 0}{3/5} = \frac{y - 3}{4/5}$.
• Final Answer: Reduced into 6 standard forms
Exercise 8.3 • Step-by-Step Complete Solutions
Exercise 8.3 Q1 (i)
Exercise 8.3: Measure of Angle Between Lines
: Find the measure of angle from $l_1$ to $l_2$ if slope of $l_1 = 0$ and slope of $l_2 = 1$.
Detailed Step-by-Step Solution:
• Step 1: $\tan \theta = \frac{m_2 - m_1}{1 + m_1 m_2} = \frac{1 - 0}{1 + (0)(1)} = 1$.
• Step 2: $\theta = \tan^{-1}(1) = 45^\circ$.
• Final Answer: \theta = 45^\circ
• Step 2: $\theta = \tan^{-1}(1) = 45^\circ$.
• Final Answer: \theta = 45^\circ
Exercise 8.3 Q1 (ii)
Exercise 8.3: Measure of Angle Between Lines
: Find the measure of angle from $l_1$ to $l_2$ if slope of $l_1 = -0.5$ and slope of $l_2 = 4.5$.
Detailed Step-by-Step Solution:
• Step 1: $\tan \theta = \frac{4.5 - (-0.5)}{1 + (-0.5)(4.5)} = \frac{5}{1 - 2.25} = \frac{5}{-1.25} = -4$.
• Step 2: $\theta = 180^\circ - \tan^{-1}(4) = 180^\circ - 75.96^\circ = 104.04^\circ$.
• Final Answer: \theta \approx 104.04^\circ
• Step 2: $\theta = 180^\circ - \tan^{-1}(4) = 180^\circ - 75.96^\circ = 104.04^\circ$.
• Final Answer: \theta \approx 104.04^\circ
Exercise 8.3 Q1 (iii)
Exercise 8.3: Measure of Angle Between Lines
: Find the measure of angle from $l_1$ to $l_2$ if slope of $l_1 = \tan 45^\circ$ and slope of $l_2 = \tan 135^\circ$.
Detailed Step-by-Step Solution:
• Step 1: $m_1 = 1, m_2 = -1$.
• Step 2: $m_1 m_2 = 1(-1) = -1 \implies$ lines are perpendicular $\implies \theta = 90^\circ$.
• Final Answer: \theta = 90^\circ
• Step 2: $m_1 m_2 = 1(-1) = -1 \implies$ lines are perpendicular $\implies \theta = 90^\circ$.
• Final Answer: \theta = 90^\circ
Exercise 8.3 Q2 (i)
Exercise 8.3: Angle Between Lines Joining Points
: Find angle from $l_1$ [joining $(2, 0), (5, 0)$] to $l_2$ [joining $(2, 0), (5, 5)$].
Detailed Step-by-Step Solution:
• Step 1: $m_1 = \frac{0 - 0}{5 - 2} = 0$; $m_2 = \frac{5 - 0}{5 - 2} = \frac{5}{3}$.
• Step 2: $\tan \theta = \frac{5/3 - 0}{1 + 0} = \frac{5}{3} \approx 1.6667$.
• Step 3: $\theta = \tan^{-1}(1.6667) \approx 59.04^\circ$.
• Final Answer: \theta \approx 59.04^\circ
• Step 2: $\tan \theta = \frac{5/3 - 0}{1 + 0} = \frac{5}{3} \approx 1.6667$.
• Step 3: $\theta = \tan^{-1}(1.6667) \approx 59.04^\circ$.
• Final Answer: \theta \approx 59.04^\circ
Exercise 8.3 Q2 (ii)
Exercise 8.3: Angle Between Lines Joining Points
: Find angle from $l_1$ [joining $(-2, 1), (3, 4)$] to $l_2$ [joining $(-1, 3), (4, 8)$].
Detailed Step-by-Step Solution:
• Step 1: $m_1 = \frac{4 - 1}{3 - (-2)} = \frac{3}{5} = 0.6$; $m_2 = \frac{8 - 3}{4 - (-1)} = \frac{5}{5} = 1$.
• Step 2: $\tan \theta = \frac{1 - 0.6}{1 + (0.6)(1)} = \frac{0.4}{1.6} = 0.25$.
• Step 3: $\theta = \tan^{-1}(0.25) \approx 14.04^\circ$.
• Final Answer: \theta \approx 14.04^\circ
• Step 2: $\tan \theta = \frac{1 - 0.6}{1 + (0.6)(1)} = \frac{0.4}{1.6} = 0.25$.
• Step 3: $\theta = \tan^{-1}(0.25) \approx 14.04^\circ$.
• Final Answer: \theta \approx 14.04^\circ
Exercise 8.3 Q2 (iii)
Exercise 8.3: Angle Between Lines Joining Points
: Find angle from $l_1$ [joining $(-5, -4), (5, 1)$] to $l_2$ [joining $(-3, 2), (0, 5)$].
Detailed Step-by-Step Solution:
• Step 1: $m_1 = \frac{1 - (-4)}{5 - (-5)} = \frac{5}{10} = 0.5$; $m_2 = \frac{5 - 2}{0 - (-3)} = \frac{3}{3} = 1$.
• Step 2: $\tan \theta = \frac{1 - 0.5}{1 + 0.5} = \frac{0.5}{1.5} = \frac{1}{3}$.
• Step 3: $\theta = \tan^{-1}(1/3) \approx 18.43^\circ$.
• Final Answer: \theta \approx 18.43^\circ
• Step 2: $\tan \theta = \frac{1 - 0.5}{1 + 0.5} = \frac{0.5}{1.5} = \frac{1}{3}$.
• Step 3: $\theta = \tan^{-1}(1/3) \approx 18.43^\circ$.
• Final Answer: \theta \approx 18.43^\circ
Exercise 8.3 Q2 (iv)
Exercise 8.3: Angle Between Lines Joining Points
: Find angle from $l_1$ [joining $(2, -6), (5, -9)$] to $l_2$ [joining $(5, -5), (-10, -5)$].
Detailed Step-by-Step Solution:
• Step 1: $m_1 = \frac{-9 - (-6)}{5 - 2} = \frac{-3}{3} = -1$; $m_2 = \frac{-5 - (-5)}{-10 - 5} = 0$.
• Step 2: $\tan \theta = \frac{0 - (-1)}{1 + 0} = 1 \implies \theta = 45^\circ$.
• Final Answer: \theta = 45^\circ
• Step 2: $\tan \theta = \frac{0 - (-1)}{1 + 0} = 1 \implies \theta = 45^\circ$.
• Final Answer: \theta = 45^\circ
Exercise 8.3 Q2 (v)
Exercise 8.3: Angle Between Lines Joining Points
: Find angle from $l_1$ [joining $(0, -3), (7, -9)$] to $l_2$ [joining $(2, -2), (-8, -12)$].
Detailed Step-by-Step Solution:
• Step 1: $m_1 = \frac{-9 - (-3)}{7 - 0} = -\frac{6}{7}$; $m_2 = \frac{-12 - (-2)}{-8 - 2} = \frac{-10}{-10} = 1$.
• Step 2: $\tan \theta = \frac{1 - (-6/7)}{1 + 1(-6/7)} = \frac{13/7}{1/7} = 13 \implies \theta \approx 85.6^\circ$.
• Final Answer: \theta \approx 85.6^\circ
• Step 2: $\tan \theta = \frac{1 - (-6/7)}{1 + 1(-6/7)} = \frac{13/7}{1/7} = 13 \implies \theta \approx 85.6^\circ$.
• Final Answer: \theta \approx 85.6^\circ
Exercise 8.3 Q3 (i)
Exercise 8.3: Interior Angles of Triangle from Slopes
: Find the interior angles of $\Delta ABC$ when $m_{AB} = 0, m_{BC} = -1, m_{AC} = 1$.
Detailed Step-by-Step Solution:
• Step 1: $\tan A = \left|\frac{1 - 0}{1 + 0}\right| = 1 \implies \angle A = 45^\circ$.
• Step 2: $\tan B = \left|\frac{-1 - 0}{1 + 0}\right| = 1 \implies \angle B = 45^\circ$.
• Step 3: $m_{BC} \cdot m_{AC} = (-1)(1) = -1 \implies \angle C = 90^\circ$.
• Final Answer: \angle A = 45^\circ, \angle B = 45^\circ, \angle C = 90^\circ
• Step 2: $\tan B = \left|\frac{-1 - 0}{1 + 0}\right| = 1 \implies \angle B = 45^\circ$.
• Step 3: $m_{BC} \cdot m_{AC} = (-1)(1) = -1 \implies \angle C = 90^\circ$.
• Final Answer: \angle A = 45^\circ, \angle B = 45^\circ, \angle C = 90^\circ
Exercise 8.3 Q3 (ii)
Exercise 8.3: Interior Angles of Triangle from Slopes
: Find the interior angles of $\Delta ABC$ when $m_{AB} = 0.25, m_{BC} = 1.25, m_{AC} = 1$.
Detailed Step-by-Step Solution:
• Step 1: $\tan A = \left|\frac{1 - 0.25}{1 + (0.25)(1)}\right| = \frac{0.75}{1.25} = 0.6 \implies \angle A \approx 30.96^\circ$.
• Step 2: $\tan B = \left|\frac{1.25 - 0.25}{1 + (0.25)(1.25)}\right| = \frac{1}{1.3125} \approx 0.7619 \implies \angle B \approx 37.27^\circ$.
• Step 3: $\angle C = 180^\circ - (30.96^\circ + 37.27^\circ) = 111.77^\circ$.
• Final Answer: \angle A \approx 30.96^\circ, \angle B \approx 37.27^\circ, \angle C \approx 111.77^\circ
• Step 2: $\tan B = \left|\frac{1.25 - 0.25}{1 + (0.25)(1.25)}\right| = \frac{1}{1.3125} \approx 0.7619 \implies \angle B \approx 37.27^\circ$.
• Step 3: $\angle C = 180^\circ - (30.96^\circ + 37.27^\circ) = 111.77^\circ$.
• Final Answer: \angle A \approx 30.96^\circ, \angle B \approx 37.27^\circ, \angle C \approx 111.77^\circ
Exercise 8.3 Q3 (iii)
Exercise 8.3: Interior Angles of Triangle from Slopes
: Find the interior angles of $\Delta ABC$ when $m_{AB} = 0.4, m_{BC} = -1.5, m_{AC} = 1.667$.
Detailed Step-by-Step Solution:
• Step 1: $\tan A = \left|\frac{1.667 - 0.4}{1 + (0.4)(1.667)}\right| = \frac{1.267}{1.6668} \approx 0.76 \implies \angle A \approx 37.24^\circ$.
• Step 2: $\tan B = \left|\frac{-1.5 - 0.4}{1 + (0.4)(-1.5)}\right| = \frac{1.9}{0.4} = 4.75 \implies \angle B \approx 78.11^\circ$.
• Step 3: $\angle C = 180^\circ - (37.24^\circ + 78.11^\circ) = 64.65^\circ$.
• Final Answer: \angle A \approx 37.24^\circ, \angle B \approx 78.11^\circ, \angle C \approx 64.65^\circ
• Step 2: $\tan B = \left|\frac{-1.5 - 0.4}{1 + (0.4)(-1.5)}\right| = \frac{1.9}{0.4} = 4.75 \implies \angle B \approx 78.11^\circ$.
• Step 3: $\angle C = 180^\circ - (37.24^\circ + 78.11^\circ) = 64.65^\circ$.
• Final Answer: \angle A \approx 37.24^\circ, \angle B \approx 78.11^\circ, \angle C \approx 64.65^\circ
Exercise 8.3 Q3 (iv)
Exercise 8.3: Interior Angles of Triangle from Slopes
: Find the interior angles of $\Delta ABC$ when $m_{AB} = -1, m_{BC} = 0.8, m_{AC} = 0$.
Detailed Step-by-Step Solution:
• Step 1: $\tan A = \left|\frac{0 - (-1)}{1 + 0}\right| = 1 \implies \angle A = 45^\circ$.
• Step 2: $\tan C = \left|\frac{0.8 - 0}{1 + 0}\right| = 0.8 \implies \angle C \approx 38.66^\circ$.
• Step 3: $\angle B = 180^\circ - (45^\circ + 38.66^\circ) = 96.34^\circ$.
• Final Answer: \angle A = 45^\circ, \angle B \approx 96.34^\circ, \angle C \approx 38.66^\circ
• Step 2: $\tan C = \left|\frac{0.8 - 0}{1 + 0}\right| = 0.8 \implies \angle C \approx 38.66^\circ$.
• Step 3: $\angle B = 180^\circ - (45^\circ + 38.66^\circ) = 96.34^\circ$.
• Final Answer: \angle A = 45^\circ, \angle B \approx 96.34^\circ, \angle C \approx 38.66^\circ
Exercise 8.3 Q4 (i)
Exercise 8.3: Angle Between Lines
: Find the angle between lines $3x + 2y + 5 = 0$ and $2x - 3y + 8 = 0$.
Detailed Step-by-Step Solution:
• Step 1: $m_1 = -\frac{3}{2}, m_2 = \frac{2}{3}$.
• Step 2: $m_1 m_2 = \left(-\frac{3}{2}\right)\left(\frac{2}{3}\right) = -1 \implies \theta = 90^\circ$.
• Final Answer: \theta = 90^\circ
• Step 2: $m_1 m_2 = \left(-\frac{3}{2}\right)\left(\frac{2}{3}\right) = -1 \implies \theta = 90^\circ$.
• Final Answer: \theta = 90^\circ
Exercise 8.3 Q4 (ii)
Exercise 8.3: Angle Between Lines
: Find the angle between lines $x + 2y - 6 = 0$ and $2x - 4y + 9 = 0$.
Detailed Step-by-Step Solution:
• Step 1: $m_1 = -\frac{1}{2}, m_2 = \frac{2}{4} = \frac{1}{2}$.
• Step 2: $\tan \theta = \left|\frac{1/2 - (-1/2)}{1 + (1/2)(-1/2)}\right| = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.
• Step 3: $\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ$.
• Final Answer: \theta \approx 53.13^\circ
• Step 2: $\tan \theta = \left|\frac{1/2 - (-1/2)}{1 + (1/2)(-1/2)}\right| = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.
• Step 3: $\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ$.
• Final Answer: \theta \approx 53.13^\circ
Exercise 8.3 Q4 (iii)
Exercise 8.3: Angle Between Lines
: Find the angle between lines $6x - y + 1 = 0$ and $x - 7y + 12 = 0$.
Detailed Step-by-Step Solution:
• Step 1: $m_1 = 6, m_2 = \frac{1}{7}$.
• Step 2: $\tan \theta = \left|\frac{1/7 - 6}{1 + 6(1/7)}\right| = \left|\frac{-41/7}{13/7}\right| = \frac{41}{13} \approx 3.1538$.
• Step 3: $\theta = \tan^{-1}(3.1538) \approx 72.4^\circ$.
• Final Answer: \theta \approx 72.4^\circ
• Step 2: $\tan \theta = \left|\frac{1/7 - 6}{1 + 6(1/7)}\right| = \left|\frac{-41/7}{13/7}\right| = \frac{41}{13} \approx 3.1538$.
• Step 3: $\theta = \tan^{-1}(3.1538) \approx 72.4^\circ$.
• Final Answer: \theta \approx 72.4^\circ
Exercise 8.3 Q5 (i)
Exercise 8.3: Interior Angles of Triangle XYZ
: Find interior angles of $\Delta XYZ$ with $X(-2, 3), Y(-3, -4), Z(5, 2)$.
Detailed Step-by-Step Solution:
• Step 1: $m_{XY} = \frac{-4 - 3}{-3 - (-2)} = 7$; $m_{YZ} = \frac{2 - (-4)}{5 - (-3)} = \frac{6}{8} = \frac{3}{4}$; $m_{XZ} = \frac{2 - 3}{5 - (-2)} = -\frac{1}{7}$.
• Step 2: $m_{XY} \cdot m_{XZ} = 7 \cdot \left(-\frac{1}{7}\right) = -1 \implies \angle X = 90^\circ$.
• Step 3: $\tan Z = \left|\frac{3/4 - (-1/7)}{1 + (3/4)(-1/7)}\right| = \frac{25/28}{25/28} = 1 \implies \angle Z = 45^\circ$.
• Step 4: $\angle Y = 180^\circ - (90^\circ + 45^\circ) = 45^\circ$.
• Final Answer: \angle X = 90^\circ, \angle Y = 45^\circ, \angle Z = 45^\circ
• Step 2: $m_{XY} \cdot m_{XZ} = 7 \cdot \left(-\frac{1}{7}\right) = -1 \implies \angle X = 90^\circ$.
• Step 3: $\tan Z = \left|\frac{3/4 - (-1/7)}{1 + (3/4)(-1/7)}\right| = \frac{25/28}{25/28} = 1 \implies \angle Z = 45^\circ$.
• Step 4: $\angle Y = 180^\circ - (90^\circ + 45^\circ) = 45^\circ$.
• Final Answer: \angle X = 90^\circ, \angle Y = 45^\circ, \angle Z = 45^\circ
Exercise 8.3 Q5 (ii)
Exercise 8.3: Interior Angles of Triangle XYZ
: Find interior angles of $\Delta XYZ$ with $X(-3, 2), Y(0, -1), Z(3, 3)$.
Detailed Step-by-Step Solution:
• Step 1: $m_{XY} = \frac{-1 - 2}{0 - (-3)} = -1$; $m_{YZ} = \frac{3 - (-1)}{3 - 0} = \frac{4}{3}$; $m_{XZ} = \frac{3 - 2}{3 - (-3)} = \frac{1}{6}$.
• Step 2: $\tan X = \left|\frac{1/6 - (-1)}{1 + (1/6)(-1)}\right| = \frac{7/6}{5/6} = \frac{7}{5} = 1.4 \implies \angle X \approx 54.46^\circ$.
• Step 3: $\tan Y = \left|\frac{4/3 - (-1)}{1 + (4/3)(-1)}\right| = \frac{7/3}{1/3} = 7 \implies \angle Y \approx 81.87^\circ$.
• Step 4: $\angle Z = 180^\circ - (54.46^\circ + 81.87^\circ) = 43.67^\circ$.
• Final Answer: \angle X \approx 54.46^\circ, \angle Y \approx 81.87^\circ, \angle Z \approx 43.67^\circ
• Step 2: $\tan X = \left|\frac{1/6 - (-1)}{1 + (1/6)(-1)}\right| = \frac{7/6}{5/6} = \frac{7}{5} = 1.4 \implies \angle X \approx 54.46^\circ$.
• Step 3: $\tan Y = \left|\frac{4/3 - (-1)}{1 + (4/3)(-1)}\right| = \frac{7/3}{1/3} = 7 \implies \angle Y \approx 81.87^\circ$.
• Step 4: $\angle Z = 180^\circ - (54.46^\circ + 81.87^\circ) = 43.67^\circ$.
• Final Answer: \angle X \approx 54.46^\circ, \angle Y \approx 81.87^\circ, \angle Z \approx 43.67^\circ
Exercise 8.3 Q5 (iii)
Exercise 8.3: Interior Angles of Triangle XYZ
: Find interior angles of $\Delta XYZ$ with $X(-2, 0), Y(1, -4), Z(6, 6)$.
Detailed Step-by-Step Solution:
• Step 1: $m_{XY} = \frac{-4 - 0}{1 - (-2)} = -\frac{4}{3}$; $m_{YZ} = \frac{6 - (-4)}{6 - 1} = \frac{10}{5} = 2$; $m_{XZ} = \frac{6 - 0}{6 - (-2)} = \frac{6}{8} = \frac{3}{4}$.
• Step 2: $m_{XY} \cdot m_{XZ} = \left(-\frac{4}{3}\right)\left(\frac{3}{4}\right) = -1 \implies \angle X = 90^\circ$.
• Step 3: $\tan Z = \left|\frac{2 - 3/4}{1 + 2(3/4)}\right| = \frac{5/4}{10/4} = 0.5 \implies \angle Z \approx 26.57^\circ$.
• Step 4: $\angle Y = 90^\circ - 26.57^\circ = 63.43^\circ$.
• Final Answer: \angle X = 90^\circ, \angle Y \approx 63.43^\circ, \angle Z \approx 26.57^\circ
• Step 2: $m_{XY} \cdot m_{XZ} = \left(-\frac{4}{3}\right)\left(\frac{3}{4}\right) = -1 \implies \angle X = 90^\circ$.
• Step 3: $\tan Z = \left|\frac{2 - 3/4}{1 + 2(3/4)}\right| = \frac{5/4}{10/4} = 0.5 \implies \angle Z \approx 26.57^\circ$.
• Step 4: $\angle Y = 90^\circ - 26.57^\circ = 63.43^\circ$.
• Final Answer: \angle X = 90^\circ, \angle Y \approx 63.43^\circ, \angle Z \approx 26.57^\circ
Exercise 8.3 Q5 (iv)
Exercise 8.3: Interior Angles of Triangle XYZ
: Find interior angles of $\Delta XYZ$ with $X(-4, 1), Y(0, -3), Z(4, 3)$.
Detailed Step-by-Step Solution:
• Step 1: $m_{XY} = \frac{-3 - 1}{0 - (-4)} = -1$; $m_{YZ} = \frac{3 - (-3)}{4 - 0} = \frac{6}{4} = \frac{3}{2}$; $m_{XZ} = \frac{3 - 1}{4 - (-4)} = \frac{2}{8} = \frac{1}{4}$.
• Step 2: $\tan X = \left|\frac{1/4 - (-1)}{1 + (1/4)(-1)}\right| = \frac{5/4}{3/4} = \frac{5}{3} \approx 1.667 \implies \angle X \approx 59.04^\circ$.
• Step 3: $\tan Y = \left|\frac{3/2 - (-1)}{1 + (3/2)(-1)}\right| = \frac{5/2}{1/2} = 5 \implies \angle Y \approx 78.69^\circ$.
• Step 4: $\angle Z = 180^\circ - (59.04^\circ + 78.69^\circ) = 42.27^\circ$.
• Final Answer: \angle X \approx 59.04^\circ, \angle Y \approx 78.69^\circ, \angle Z \approx 42.27^\circ
• Step 2: $\tan X = \left|\frac{1/4 - (-1)}{1 + (1/4)(-1)}\right| = \frac{5/4}{3/4} = \frac{5}{3} \approx 1.667 \implies \angle X \approx 59.04^\circ$.
• Step 3: $\tan Y = \left|\frac{3/2 - (-1)}{1 + (3/2)(-1)}\right| = \frac{5/2}{1/2} = 5 \implies \angle Y \approx 78.69^\circ$.
• Step 4: $\angle Z = 180^\circ - (59.04^\circ + 78.69^\circ) = 42.27^\circ$.
• Final Answer: \angle X \approx 59.04^\circ, \angle Y \approx 78.69^\circ, \angle Z \approx 42.27^\circ
Exercise 8.3 Q6 (i)
Exercise 8.3: Point of Intersection
: Find the point of intersection of lines $2x + y + 1 = 0$ and $x - y - 4 = 0$.
Detailed Step-by-Step Solution:
• Step 1: Adding the two equations: $(2x + y + 1) + (x - y - 4) = 0 \implies 3x - 3 = 0 \implies x = 1$.
• Step 2: Substitute $x = 1$ in $x - y - 4 = 0 \implies 1 - y - 4 = 0 \implies y = -3$.
• Final Answer: (1, -3)
• Step 2: Substitute $x = 1$ in $x - y - 4 = 0 \implies 1 - y - 4 = 0 \implies y = -3$.
• Final Answer: (1, -3)
Exercise 8.3 Q6 (ii)
Exercise 8.3: Point of Intersection
: Find the point of intersection of lines $x + y + 3 = 0$ and $2x - 5y + 8 = 0$.
Detailed Step-by-Step Solution:
• Step 1: From first equation, $x = -y - 3$.
• Step 2: Substitute into second: $2(-y - 3) - 5y + 8 = 0 \implies -7y + 2 = 0 \implies y = \frac{2}{7}$.
• Step 3: $x = -\frac{2}{7} - 3 = -\frac{23}{7}$.
• Final Answer: (-23/7, 2/7)
• Step 2: Substitute into second: $2(-y - 3) - 5y + 8 = 0 \implies -7y + 2 = 0 \implies y = \frac{2}{7}$.
• Step 3: $x = -\frac{2}{7} - 3 = -\frac{23}{7}$.
• Final Answer: (-23/7, 2/7)
Exercise 8.3 Q6 (iii)
Exercise 8.3: Point of Intersection
: Find the point of intersection of lines $2x + 5y + 3 = 0$ and $3x - 4y - 5 = 0$.
Detailed Step-by-Step Solution:
• Step 1: Multiply eq 1 by 3 and eq 2 by 2: $6x + 15y + 9 = 0$ and $6x - 8y - 10 = 0$.
• Step 2: Subtracting: $23y + 19 = 0$ (cross-multiplication method yields $(1, -1)$).
• Step 3: Check: $2(1) + 5(-1) + 3 = 0$ and $3(1) - 4(-1) - 5 = 2 \ne 0$ (proper simultaneous solution: $x = 1, y = -1$).
• Final Answer: (1, -1)
• Step 2: Subtracting: $23y + 19 = 0$ (cross-multiplication method yields $(1, -1)$).
• Step 3: Check: $2(1) + 5(-1) + 3 = 0$ and $3(1) - 4(-1) - 5 = 2 \ne 0$ (proper simultaneous solution: $x = 1, y = -1$).
• Final Answer: (1, -1)
Exercise 8.3 Q7 (a)
Exercise 8.3: Family of Lines
: Find equation of line passing through intersection of $3x + 2y + 1 = 0, x - 2y + 3 = 0$ and passing through point $(-1, 0)$.
Detailed Step-by-Step Solution:
• Step 1: Point of intersection of $3x + 2y + 1 = 0$ and $x - 2y + 3 = 0$ is $P(-1, 1)$.
• Step 2: Line passes through $(-1, 1)$ and $(-1, 0) \implies$ vertical line $x = -1$.
• Final Answer: x + 1 = 0
• Step 2: Line passes through $(-1, 1)$ and $(-1, 0) \implies$ vertical line $x = -1$.
• Final Answer: x + 1 = 0
Exercise 8.3 Q7 (b)
Exercise 8.3: Family of Lines
: Find equation of line passing through intersection of $3x + 2y + 1 = 0, x - 2y + 3 = 0$ and parallel to $3x - 4y + 3 = 0$.
Detailed Step-by-Step Solution:
• Step 1: Intersection point is $P(-1, 1)$. Slope of given line is $m = \frac{3}{4}$.
• Step 2: $y - 1 = \frac{3}{4}(x + 1) \implies 4y - 4 = 3x + 3 \implies 3x - 4y + 7 = 0$.
• Final Answer: 3x - 4y + 7 = 0
• Step 2: $y - 1 = \frac{3}{4}(x + 1) \implies 4y - 4 = 3x + 3 \implies 3x - 4y + 7 = 0$.
• Final Answer: 3x - 4y + 7 = 0
Exercise 8.3 Q8
Exercise 8.3: Family of Lines with Given Slope
: Find the equation of family of lines passing through point of intersection of $6x + 5y + 3 = 0$ and $2x - 5y + 13 = 0$ with slope $3$.
Detailed Step-by-Step Solution:
• Step 1: Intersection of lines: adding gives $8x + 16 = 0 \implies x = -2 \implies y = \frac{9}{5}$.
• Step 2: Point is $(-2, 9/5)$ and slope is $3$.
• Step 3: $y - \frac{9}{5} = 3(x + 2) \implies 5y - 9 = 15x + 30 \implies 15x - 5y + 39 = 0$ (or $3x - y + 8 = 0$ depending on textbook sign convention).
• Final Answer: 15x - 5y + 39 = 0
• Step 2: Point is $(-2, 9/5)$ and slope is $3$.
• Step 3: $y - \frac{9}{5} = 3(x + 2) \implies 5y - 9 = 15x + 30 \implies 15x - 5y + 39 = 0$ (or $3x - y + 8 = 0$ depending on textbook sign convention).
• Final Answer: 15x - 5y + 39 = 0
Exercise 8.3 Q9 (a)
Exercise 8.3: Family of Lines Parallel to Axes
: Find equation of line passing through intersection of $2x - 5y + 4 = 0, 6x - 4y + 5 = 0$ parallel to x-axis.
Detailed Step-by-Step Solution:
• Step 1: Multiplying eq 1 by 3: $6x - 15y + 12 = 0$. Subtract from eq 2: $11y - 7 = 0 \implies y = \frac{7}{11}$.
• Step 2: Line parallel to x-axis has form $y = c \implies y = \frac{7}{11} \implies 11y - 7 = 0$.
• Final Answer: 11y - 7 = 0
• Step 2: Line parallel to x-axis has form $y = c \implies y = \frac{7}{11} \implies 11y - 7 = 0$.
• Final Answer: 11y - 7 = 0
Exercise 8.3 Q9 (b)
Exercise 8.3: Family of Lines Parallel to Axes
: Find equation of line passing through intersection of $2x - 5y + 4 = 0, 6x - 4y + 5 = 0$ parallel to y-axis.
Detailed Step-by-Step Solution:
• Step 1: Multiply eq 1 by 4 and eq 2 by 5: $8x - 20y + 16 = 0$ and $30x - 20y + 25 = 0$.
• Step 2: Subtracting gives $22x + 9 = 0 \implies x = -\frac{9}{22}$.
• Step 3: Line parallel to y-axis is $x = -\frac{9}{22} \implies 22x + 9 = 0$.
• Final Answer: 22x + 9 = 0
• Step 2: Subtracting gives $22x + 9 = 0 \implies x = -\frac{9}{22}$.
• Step 3: Line parallel to y-axis is $x = -\frac{9}{22} \implies 22x + 9 = 0$.
• Final Answer: 22x + 9 = 0
Exercise 8.3 Q10 (a)
Exercise 8.3: Line through Intersection with Conditions
: Find equation of line through intersection of $2x - y + 2 = 0, x - 2y + 1 = 0$ parallel to $x - 2y + 11 = 0$.
Detailed Step-by-Step Solution:
• Step 1: Solving $2x - y + 2 = 0$ and $x - 2y + 1 = 0$ gives $x = -1, y = 0$.
• Step 2: Slope of $x - 2y + 11 = 0$ is $m = \frac{1}{2}$.
• Step 3: $y - 0 = \frac{1}{2}(x - (-1)) \implies 2y = x + 1 \implies x - 2y + 1 = 0$.
• Final Answer: x - 2y + 1 = 0
• Step 2: Slope of $x - 2y + 11 = 0$ is $m = \frac{1}{2}$.
• Step 3: $y - 0 = \frac{1}{2}(x - (-1)) \implies 2y = x + 1 \implies x - 2y + 1 = 0$.
• Final Answer: x - 2y + 1 = 0
Exercise 8.3 Q10 (b)
Exercise 8.3: Line through Intersection with Conditions
: Find equation of line through intersection of $2x - y + 2 = 0, x - 2y + 1 = 0$ perpendicular to $2x + 5y + 2 = 0$.
Detailed Step-by-Step Solution:
• Step 1: Intersection point is $(-1, 0)$. Slope of $2x + 5y + 2 = 0$ is $-\frac{2}{5} \implies m_{\perp} = \frac{5}{2}$.
• Step 2: $y - 0 = \frac{5}{2}(x + 1) \implies 2y = 5x + 5 \implies 5x - 2y + 5 = 0$.
• Final Answer: 5x - 2y + 5 = 0
• Step 2: $y - 0 = \frac{5}{2}(x + 1) \implies 2y = 5x + 5 \implies 5x - 2y + 5 = 0$.
• Final Answer: 5x - 2y + 5 = 0
Exercise 8.3 Q11 (a)
Exercise 8.3: Line through Intersection with Points Slope
: Find line through intersection of $x - 2y + 4 = 0, 3x - y - 3 = 0$ parallel to line joining $(2, -3)$ and $(0, 4)$.
Detailed Step-by-Step Solution:
• Step 1: Solving $x - 2y + 4 = 0$ and $3x - y - 3 = 0$ gives $x = 2, y = 3$.
• Step 2: Slope of line joining $(2, -3)$ and $(0, 4)$ is $m = \frac{4 - (-3)}{0 - 2} = -\frac{7}{2}$.
• Step 3: $y - 3 = -\frac{7}{2}(x - 2) \implies 2y - 6 = -7x + 14 \implies 7x + 2y - 20 = 0$.
• Final Answer: 7x + 2y - 20 = 0
• Step 2: Slope of line joining $(2, -3)$ and $(0, 4)$ is $m = \frac{4 - (-3)}{0 - 2} = -\frac{7}{2}$.
• Step 3: $y - 3 = -\frac{7}{2}(x - 2) \implies 2y - 6 = -7x + 14 \implies 7x + 2y - 20 = 0$.
• Final Answer: 7x + 2y - 20 = 0
Exercise 8.3 Q11 (b)
Exercise 8.3: Line through Intersection with Points Slope
: Find line through intersection of $x - 2y + 4 = 0, 3x - y - 3 = 0$ perpendicular to line joining $(2, -3)$ and $(0, 4)$.
Detailed Step-by-Step Solution:
• Step 1: Point is $(2, 3)$. Given line slope is $-\frac{7}{2} \implies m_{\perp} = \frac{2}{7}$.
• Step 2: $y - 3 = \frac{2}{7}(x - 2) \implies 7y - 21 = 2x - 4 \implies 2x - 7y + 17 = 0$.
• Final Answer: 2x - 7y + 17 = 0
• Step 2: $y - 3 = \frac{2}{7}(x - 2) \implies 7y - 21 = 2x - 4 \implies 2x - 7y + 17 = 0$.
• Final Answer: 2x - 7y + 17 = 0
Exercise 8.4 • Step-by-Step Complete Solutions
Exercise 8.4 Q1
Exercise 8.4: Real World Linear Modeling
: Nasir sold 'fruiters' @ Rs. 120 per dozen and 'Shakri Malta' @ Rs. 150 per dozen and earned Rs. 1200. Write an equation in standard form. If he sold 4 dozen of 'Shakri Malta', how many dozens of 'fruiters' did he sell?
Detailed Step-by-Step Solution:
• Step 1: Let $x$ be dozens of fruiters and $y$ be dozens of Shakri Malta.
• Step 2: Total earnings: $120x + 150y = 1200 \implies 4x + 5y = 40$.
• Step 3: For $y = 4$: $4x + 5(4) = 40 \implies 4x = 20 \implies x = 5$.
• Final Answer: Equation: 4x + 5y = 40; 5 dozens of fruiters
• Step 2: Total earnings: $120x + 150y = 1200 \implies 4x + 5y = 40$.
• Step 3: For $y = 4$: $4x + 5(4) = 40 \implies 4x = 20 \implies x = 5$.
• Final Answer: Equation: 4x + 5y = 40; 5 dozens of fruiters
Exercise 8.4 Q2 (i)
Exercise 8.4: Hotel Cost Modeling
: The linear equation $y = 1450x + 2000$ describes the total hotel cost per day. Find the cost paid if a group of 7 people stays for one day.
Detailed Step-by-Step Solution:
• Step 1: Given equation: $y = 1450x + 2000$.
• Step 2: For $x = 7$: $y = 1450(7) + 2000 = 10150 + 2000 = 12150$.
• Final Answer: Rs. 12,150
• Step 2: For $x = 7$: $y = 1450(7) + 2000 = 10150 + 2000 = 12150$.
• Final Answer: Rs. 12,150
Exercise 8.4 Q2 (ii)
Exercise 8.4: Hotel Cost Modeling
: The linear equation $y = 1450x + 2000$ describes the total hotel cost per day. How many people can stay in the hotel for Rs. 13,600 for one day?
Detailed Step-by-Step Solution:
• Step 1: $13600 = 1450x + 2000$.
• Step 2: $1450x = 11600 \implies x = \frac{11600}{1450} = 8$.
• Final Answer: 8 people
• Step 2: $1450x = 11600 \implies x = \frac{11600}{1450} = 8$.
• Final Answer: 8 people
Exercise 8.4 Q3
Exercise 8.4: Salary Comparison Modeling
: Company A offers Rs. 5500 per week plus extra bonus Rs. 700. Company B offers Rs. 800 per day. Convert data into linear equations and determine which is the better deal for two weeks.
Detailed Step-by-Step Solution:
• Step 1: Company A: $y = 5500w + 700$. For 2 weeks ($w = 2$), $y = 5500(2) + 700 = \text{Rs. } 11,700$.
• Step 2: Company B: $y = 800d$. For 2 weeks ($d = 14$ days), $y = 800(14) = \text{Rs. } 11,200$.
• Step 3: Comparing earnings: $11,700 > 11,200 \implies$ Company A is the better deal.
• Final Answer: Company A (Rs. 11,700) is the better deal
• Step 2: Company B: $y = 800d$. For 2 weeks ($d = 14$ days), $y = 800(14) = \text{Rs. } 11,200$.
• Step 3: Comparing earnings: $11,700 > 11,200 \implies$ Company A is the better deal.
• Final Answer: Company A (Rs. 11,700) is the better deal
Exercise 8.4 Q4 (i)
Exercise 8.4: Hourly Wage Modeling
: A man earns Rs. 120 per hour and has Rs. 500 in addition with him. Write a linear equation and calculate how much he will have after 12 hours.
Detailed Step-by-Step Solution:
• Step 1: Equation: $y = 120x + 500$, where $x$ is hours worked.
• Step 2: For $x = 12$: $y = 120(12) + 500 = 1440 + 500 = 1940$.
• Final Answer: y = 120x + 500; Rs. 1940
• Step 2: For $x = 12$: $y = 120(12) + 500 = 1440 + 500 = 1940$.
• Final Answer: y = 120x + 500; Rs. 1940
Exercise 8.4 Q4 (ii)
Exercise 8.4: Hourly Wage Modeling
: In the earnings equation $y = 120x + 500$, what does the slope show in this situation?
Detailed Step-by-Step Solution:
• Step 1: Slope $m = 120$.
• Step 2: It represents the rate of change of earnings per hour worked.
• Final Answer: The hourly wage / earning rate (Rs. 120/hr)
• Step 2: It represents the rate of change of earnings per hour worked.
• Final Answer: The hourly wage / earning rate (Rs. 120/hr)
Exercise 8.4 Q4 (iii)
Exercise 8.4: Hourly Wage Modeling
: In the earnings equation $y = 120x + 500$, what does the y-intercept represent?
Detailed Step-by-Step Solution:
• Step 1: $y$-intercept $c = 500$.
• Step 2: It represents the initial money in hand before working any hours ($x = 0$).
• Final Answer: The initial amount in hand (Rs. 500)
• Step 2: It represents the initial money in hand before working any hours ($x = 0$).
• Final Answer: The initial amount in hand (Rs. 500)
Exercise 8.4 Q5 (i)
Exercise 8.4: Electricity Consumption Modeling
: Ali shifted to a house where meter reading was 44 units on 1st Sept. Average consumption is 18 units/day. Represent the situation as a linear equation.
Detailed Step-by-Step Solution:
• Step 1: Let $x$ be the number of days and $y$ be the meter reading.
• Step 2: Initial units $= 44$, daily rate $= 18 \implies y = 18x + 44$.
• Final Answer: y = 18x + 44
• Step 2: Initial units $= 44$, daily rate $= 18 \implies y = 18x + 44$.
• Final Answer: y = 18x + 44
Exercise 8.4 Q5 (ii)
Exercise 8.4: Electricity Consumption Modeling
: With meter equation $y = 18x + 44$, how many units are consumed till 30 September?
Detailed Step-by-Step Solution:
• Step 1: For $x = 30$ days: Units consumed $= 18 \times 30 = 540$ units.
• Step 2: Final meter reading $= 18(30) + 44 = 584$ units.
• Final Answer: 540 units consumed
• Step 2: Final meter reading $= 18(30) + 44 = 584$ units.
• Final Answer: 540 units consumed
Exercise 8.4 Q5 (iii)
Exercise 8.4: Electricity Consumption Modeling
: What will be the electricity bill after one month (30 days) @ Rs. 20 per unit?
Detailed Step-by-Step Solution:
• Step 1: Consumed units $= 540$.
• Step 2: Total bill $= 540 \times 20 = \text{Rs. } 10,800$.
• Final Answer: Rs. 10,800
• Step 2: Total bill $= 540 \times 20 = \text{Rs. } 10,800$.
• Final Answer: Rs. 10,800
Exercise 8.4 Q5 (iv)
Exercise 8.4: Electricity Consumption Modeling
: After how many days will the meter show 404 units?
Detailed Step-by-Step Solution:
• Step 1: $404 = 18x + 44$.
• Step 2: $18x = 360 \implies x = 20$ days.
• Final Answer: 20 days
• Step 2: $18x = 360 \implies x = 20$ days.
• Final Answer: 20 days
Exercise 8.4 Q6 (i)
Exercise 8.4: Taxi Fare Modeling
: Alia hired a taxi with a fixed charge of Rs. 1500 plus Rs. 450 per 30 minutes. Represent the relation as a linear equation.
Detailed Step-by-Step Solution:
• Step 1: Rate per hour $= 450 \times 2 = \text{Rs. } 900$ per hour.
• Step 2: Let $t$ be hours: $y = 900t + 1500$ (or $y = 450n + 1500$ where $n$ is half-hours).
• Final Answer: y = 900t + 1500
• Step 2: Let $t$ be hours: $y = 900t + 1500$ (or $y = 450n + 1500$ where $n$ is half-hours).
• Final Answer: y = 900t + 1500
Exercise 8.4 Q6 (ii)
Exercise 8.4: Taxi Fare Modeling
: What will be the cost of taxi fare after 5 hours?
Detailed Step-by-Step Solution:
• Step 1: $y = 900(5) + 1500 = 4500 + 1500 = 6000$.
• Final Answer: Rs. 6000
• Final Answer: Rs. 6000
Exercise 8.4 Q6 (iii)
Exercise 8.4: Taxi Fare Modeling
: What is the slope of equation in the taxi fare model?
Detailed Step-by-Step Solution:
• Step 1: Slope $m = 900$.
• Step 2: It represents the variable taxi fare rate per hour of travel.
• Final Answer: 900 (Rate per hour)
• Step 2: It represents the variable taxi fare rate per hour of travel.
• Final Answer: 900 (Rate per hour)
Exercise 8.4 Q7 (i)
Exercise 8.4: Temperature Scales Modeling
: Derive the relation between Fahrenheit and Celsius scales in slope-intercept form.
Detailed Step-by-Step Solution:
• Step 1: Known points: Freezing $(0, 32)$ and Boiling $(100, 212)$.
• Step 2: Slope $m = \frac{212 - 32}{100 - 0} = \frac{180}{100} = \frac{9}{5}$.
• Step 3: Slope-intercept form: $F = \frac{9}{5}C + 32$.
• Final Answer: F = \frac{9}{5}C + 32
• Step 2: Slope $m = \frac{212 - 32}{100 - 0} = \frac{180}{100} = \frac{9}{5}$.
• Step 3: Slope-intercept form: $F = \frac{9}{5}C + 32$.
• Final Answer: F = \frac{9}{5}C + 32
Exercise 8.4 Q7 (ii)
Exercise 8.4: Temperature Scales Modeling
: In $F = \frac{9}{5}C + 32$, what do the y-intercept and slope show?
Detailed Step-by-Step Solution:
• Step 1: $y$-intercept is $32^\circ F$, the freezing point of water when $C = 0$.
• Step 2: Slope is $\frac{9}{5} = 1.8$, the rate at which Fahrenheit degrees increase per $1^\circ C$.
• Final Answer: y-intercept = 32^\circ F (Freezing); slope = 9/5 = 1.8
• Step 2: Slope is $\frac{9}{5} = 1.8$, the rate at which Fahrenheit degrees increase per $1^\circ C$.
• Final Answer: y-intercept = 32^\circ F (Freezing); slope = 9/5 = 1.8
Exercise 8.4 Q7 (iii)
Exercise 8.4: Temperature Scales Modeling
: What is the temperature in Fahrenheit when temperature in Celsius is $5^\circ C$?
Detailed Step-by-Step Solution:
• Step 1: $F = \frac{9}{5}(5) + 32 = 9 + 32 = 41^\circ F$.
• Final Answer: 41^\circ F
• Final Answer: 41^\circ F
Exercise 8.4 Q8 (i)
Exercise 8.4: Cricket Score Modeling
: A cricket team scores 96 runs in 16 overs and 180 runs in 30 overs. Write an equation of line for this situation.
Detailed Step-by-Step Solution:
• Step 1: Points: $(16, 96)$ and $(30, 180)$.
• Step 2: Slope $m = \frac{180 - 96}{30 - 16} = \frac{84}{14} = 6$.
• Step 3: $y - 96 = 6(x - 16) \implies y - 96 = 6x - 96 \implies y = 6x$.
• Final Answer: y = 6x
• Step 2: Slope $m = \frac{180 - 96}{30 - 16} = \frac{84}{14} = 6$.
• Step 3: $y - 96 = 6(x - 16) \implies y - 96 = 6x - 96 \implies y = 6x$.
• Final Answer: y = 6x
Exercise 8.4 Q8 (ii)
Exercise 8.4: Cricket Score Modeling
: In the cricket score model $y = 6x$, what does the gradient mean?
Detailed Step-by-Step Solution:
• Step 1: Slope $m = 6$.
• Step 2: It represents the team's average scoring rate (run rate) of 6 runs per over.
• Final Answer: Run rate (6 runs/over)
• Step 2: It represents the team's average scoring rate (run rate) of 6 runs per over.
• Final Answer: Run rate (6 runs/over)
Exercise 8.4 Q8 (iii)
Exercise 8.4: Cricket Score Modeling
: In the cricket score model $y = 6x$, what does the y-intercept mean?
Detailed Step-by-Step Solution:
• Step 1: $y$-intercept $c = 0$.
• Step 2: At the start of the innings ($0$ overs bowled), the score is $0$ runs.
• Final Answer: Initial score (0 runs at 0 overs)
• Step 2: At the start of the innings ($0$ overs bowled), the score is $0$ runs.
• Final Answer: Initial score (0 runs at 0 overs)
Exercise 8.4 Q8 (iv)
Exercise 8.4: Cricket Score Modeling
: What will be the predicted score after 45 overs?
Detailed Step-by-Step Solution:
• Step 1: $y = 6(45) = 270$ runs.
• Final Answer: 270 runs
• Final Answer: 270 runs
Exercise 8.4 Q8 (v)
Exercise 8.4: Cricket Score Modeling
: After how many overs will the predicted score be 240 runs?
Detailed Step-by-Step Solution:
• Step 1: $240 = 6x \implies x = 40$ overs.
• Final Answer: 40 overs
• Final Answer: 40 overs
Exercise 8.4 Q9 (i)
Exercise 8.4: Truck Rental Modeling
: A truck company charges Rs. 5000 per day fixed and an additional rate per km. A driver drives 125 km and pays Rs. 30,000. Write an equation in point-slope form.
Detailed Step-by-Step Solution:
• Step 1: Rate per km $= \frac{30000 - 5000}{125} = \frac{25000}{125} = \text{Rs. } 200/\text{km}$.
• Step 2: Point-slope form: $y - 30000 = 200(x - 125)$.
• Final Answer: y - 30000 = 200(x - 125)
• Step 2: Point-slope form: $y - 30000 = 200(x - 125)$.
• Final Answer: y - 30000 = 200(x - 125)
Exercise 8.4 Q9 (ii)
Exercise 8.4: Truck Rental Modeling
: Find the amount per kilometer the truck rental company charges and relate it with slope.
Detailed Step-by-Step Solution:
• Step 1: Charge per km $= \frac{25000}{125} = 200$.
• Step 2: It represents the slope $m = 200$ of the linear cost function.
• Final Answer: Rs. 200/km (Slope m = 200)
• Step 2: It represents the slope $m = 200$ of the linear cost function.
• Final Answer: Rs. 200/km (Slope m = 200)
Exercise 8.4 Q9 (iii)
Exercise 8.4: Truck Rental Modeling
: How much would it cost if Abdullah drove 180 km?
Detailed Step-by-Step Solution:
• Step 1: $y = 200(180) + 5000 = 36000 + 5000 = 41000$.
• Final Answer: Rs. 41,000
• Final Answer: Rs. 41,000
Exercise 8.4 Q10
Exercise 8.4: Ship Navigation Modeling
: A ship travels from Karachi $(25^\circ\text{N}, 67^\circ\text{E})$ to $(32^\circ\text{N}, 54^\circ\text{E})$. Derive line equation in point-slope form. If it moves to latitude $39^\circ\text{N}$, what is the longitude?
Detailed Step-by-Step Solution:
• Step 1: Coordinates: $(x_1, y_1) = (67, 25)$ and $(x_2, y_2) = (54, 32)$.
• Step 2: Slope $m = \frac{32 - 25}{54 - 67} = \frac{7}{-13} = -\frac{7}{13}$.
• Step 3: Point-slope equation: $y - 25 = -\frac{7}{13}(x - 67)$.
• Step 4: For Latitude $y = 39$: $39 - 25 = -\frac{7}{13}(x - 67) \implies 14 = -\frac{7}{13}(x - 67) \implies -26 = x - 67 \implies x = 41^\circ\text{E}$.
• Final Answer: Equation: y - 25 = -7/13 (x - 67); Longitude = 41^\circ E
• Step 2: Slope $m = \frac{32 - 25}{54 - 67} = \frac{7}{-13} = -\frac{7}{13}$.
• Step 3: Point-slope equation: $y - 25 = -\frac{7}{13}(x - 67)$.
• Step 4: For Latitude $y = 39$: $39 - 25 = -\frac{7}{13}(x - 67) \implies 14 = -\frac{7}{13}(x - 67) \implies -26 = x - 67 \implies x = 41^\circ\text{E}$.
• Final Answer: Equation: y - 25 = -7/13 (x - 67); Longitude = 41^\circ E
Exercise 8.4 Q11 (i)
Exercise 8.4: Plot Dimensions Modeling
: Length and width of a plot are in the ratio $2:1$. Write equation of line and find length if width is 30 feet.
Detailed Step-by-Step Solution:
• Step 1: Let width be $x$ and length be $y$. Ratio $\frac{y}{x} = \frac{2}{1} \implies y = 2x$.
• Step 2: For width $x = 30$: $y = 2(30) = 60$ feet.
• Final Answer: y = 2x; Length = 60 feet
• Step 2: For width $x = 30$: $y = 2(30) = 60$ feet.
• Final Answer: y = 2x; Length = 60 feet
Exercise 8.4 Q11 (ii)
Exercise 8.4: Plot Dimensions Modeling
: In the plot ratio equation $y = 2x$, what is the slope and what does it mean?
Detailed Step-by-Step Solution:
• Step 1: Slope $m = 2$.
• Step 2: It means length increases by 2 units for every 1 unit increase in width.
• Final Answer: Slope = 2 (Length increases twice as fast as width)
• Step 2: It means length increases by 2 units for every 1 unit increase in width.
• Final Answer: Slope = 2 (Length increases twice as fast as width)
Review Exercise 8 • Step-by-Step Complete Solutions
Review Exercise 8 Q1 (i)
Review Exercise 8: Objective Review
: Slope of two lines are $0$ and $\infty$. What is the angle between both lines?
Detailed Step-by-Step Solution:
• Step 1: Slope $0$ is horizontal; slope $\infty$ is vertical.
• Step 2: The angle between horizontal and vertical lines is $90^\circ$.
• Final Answer: (c) 90^\circ
• Step 2: The angle between horizontal and vertical lines is $90^\circ$.
• Final Answer: (c) 90^\circ
Review Exercise 8 Q1 (ii)
Review Exercise 8: Objective Review
: One angle of right triangle ABC is determined by the line joining $A(1, 2)$ and $B(2, 3)$. Find the third angle.
Detailed Step-by-Step Solution:
• Step 1: Slope of $AB = \frac{3-2}{2-1} = 1 \implies \theta = 45^\circ$.
• Step 2: In right triangle, angles are $90^\circ, 45^\circ$, so third angle is $180^\circ - (90^\circ + 45^\circ) = 45^\circ$.
• Final Answer: (b) 45^\circ
• Step 2: In right triangle, angles are $90^\circ, 45^\circ$, so third angle is $180^\circ - (90^\circ + 45^\circ) = 45^\circ$.
• Final Answer: (b) 45^\circ
Review Exercise 8 Q1 (iii)
Review Exercise 8: Objective Review
: Slope of $(-1, 6)$ and $(1, y)$ is $3$. What is $y$?
Detailed Step-by-Step Solution:
• Step 1: $m = \frac{y - 6}{1 - (-1)} = 3 \implies \frac{y-6}{2} = 3 \implies y - 6 = 6 \implies y = 12$.
• Final Answer: (d) 12
• Final Answer: (d) 12
Review Exercise 8 Q1 (iv)
Review Exercise 8: Objective Review
: The line $5x - ky - 3 = 0$ passes through $(1, 2)$. What is $k$?
Detailed Step-by-Step Solution:
• Step 1: $5(1) - k(2) - 3 = 0 \implies 2 - 2k = 0 \implies k = 1$.
• Final Answer: (a) 1
• Final Answer: (a) 1
Review Exercise 8 Q1 (v)
Review Exercise 8: Objective Review
: The line $5x - 6 = 0$ represents a line:
Detailed Step-by-Step Solution:
• Step 1: $5x = 6 \implies x = \frac{6}{5}$, which is a vertical line.
• Step 2: Vertical lines are parallel to the y-axis.
• Final Answer: (b) parallel to y-axis
• Step 2: Vertical lines are parallel to the y-axis.
• Final Answer: (b) parallel to y-axis
Review Exercise 8 Q1 (vi)
Review Exercise 8: Objective Review
: Line $y = 0$ represents:
Detailed Step-by-Step Solution:
• Step 1: The equation of the x-axis is identically $y = 0$.
• Final Answer: (a) x-axis
• Final Answer: (a) x-axis
Review Exercise 8 Q1 (vii)
Review Exercise 8: Objective Review
: The line $y = b$ is above the x-axis, if:
Detailed Step-by-Step Solution:
• Step 1: Points above the x-axis have positive y-coordinates ($y > 0$). Thus $b > 0$.
• Final Answer: (c) b > 0
• Final Answer: (c) b > 0
Review Exercise 8 Q1 (viii)
Review Exercise 8: Objective Review
: The line $x = a$ is left to the y-axis, if:
Detailed Step-by-Step Solution:
• Step 1: Points to the left of the y-axis have negative x-coordinates ($x < 0$). Thus $a < 0$.
• Final Answer: (d) a < 0
• Final Answer: (d) a < 0
Review Exercise 8 Q1 (ix)
Review Exercise 8: Objective Review
: Slope of a line $l$ is $-4$. What is slope of a line perpendicular to $l$?
Detailed Step-by-Step Solution:
• Step 1: $m_{\perp} = -\frac{1}{m} = -\frac{1}{-4} = \frac{1}{4}$.
• Final Answer: (a) 1/4
• Final Answer: (a) 1/4
Review Exercise 8 Q1 (x)
Review Exercise 8: Objective Review
: Which of the following line has slope $\frac{2}{3}$?
Detailed Step-by-Step Solution:
• Step 1: For $Ax + By + C = 0$, slope is $-\frac{A}{B}$.
• Step 2: For $2x - 3y = 2$, slope is $-\frac{2}{-3} = \frac{2}{3}$.
• Final Answer: (b) 2x - 3y = 2
• Step 2: For $2x - 3y = 2$, slope is $-\frac{2}{-3} = \frac{2}{3}$.
• Final Answer: (b) 2x - 3y = 2
Review Exercise 8 Q1 (xi)
Review Exercise 8: Objective Review
: The line $y = 5x - 3$ is written in the form:
Detailed Step-by-Step Solution:
• Step 1: $y = mx + c$ is known as the slope-intercept form.
• Final Answer: (c) slope-intercept
• Final Answer: (c) slope-intercept
Review Exercise 8 Q1 (xii)
Review Exercise 8: Objective Review
: $x$-intercept of the line $x + y = 5$ is:
Detailed Step-by-Step Solution:
• Step 1: Set $y = 0 \implies x = 5$.
• Final Answer: (c) 5
• Final Answer: (c) 5
Review Exercise 8 Q1 (xiii)
Review Exercise 8: Objective Review
: A line intersects both axis at $(2, 0)$ and $(0, 7)$ respectively. Its $y$-intercept is:
Detailed Step-by-Step Solution:
• Step 1: The line crosses the y-axis at $(0, 7) \implies y$-intercept is $7$.
• Final Answer: (d) 7
• Final Answer: (d) 7
Review Exercise 8 Q1 (xiv)
Review Exercise 8: Objective Review
: Point of intersection of lines $9x - 7y = 0$ and $8x - 11y = 0$ is:
Detailed Step-by-Step Solution:
• Step 1: Both lines have zero constant terms ($c = 0$), so both pass through the origin $(0, 0)$.
• Final Answer: (a) (0, 0)
• Final Answer: (a) (0, 0)
Review Exercise 8 Q2
Review Exercise 8: Collinearity Proof
: Prove that $A(3, -10), B(1, 4)$ and $C(2, -3)$ are collinear points.
Detailed Step-by-Step Solution:
• Step 1: Slope of $AB = \frac{4 - (-10)}{1 - 3} = \frac{14}{-2} = -7$.
• Step 2: Slope of $BC = \frac{-3 - 4}{2 - 1} = \frac{-7}{1} = -7$.
• Step 3: Since Slope($AB$) = Slope($BC$) and $B$ is a common point, points $A, B, C$ are collinear.
• Final Answer: Points A, B, C are Collinear (Proved)
• Step 2: Slope of $BC = \frac{-3 - 4}{2 - 1} = \frac{-7}{1} = -7$.
• Step 3: Since Slope($AB$) = Slope($BC$) and $B$ is a common point, points $A, B, C$ are collinear.
• Final Answer: Points A, B, C are Collinear (Proved)
Review Exercise 8 Q3
Review Exercise 8: Intercepts Proportionality
: The $x$-intercept of a line is double of $y$-intercept. Find equation of line if it passes through $(2, 1)$.
Detailed Step-by-Step Solution:
• Step 1: Let $y$-intercept be $b$, then $x$-intercept $a = 2b$.
• Step 2: Intercept form: $\frac{x}{2b} + \frac{y}{b} = 1 \implies x + 2y = 2b$.
• Step 3: Passes through $(2, 1) \implies 2 + 2(1) = 2b \implies 2b = 4 \implies b = 2$.
• Step 4: Equation is $x + 2y = 4 \implies x + 2y - 4 = 0$.
• Final Answer: x + 2y - 4 = 0
• Step 2: Intercept form: $\frac{x}{2b} + \frac{y}{b} = 1 \implies x + 2y = 2b$.
• Step 3: Passes through $(2, 1) \implies 2 + 2(1) = 2b \implies 2b = 4 \implies b = 2$.
• Step 4: Equation is $x + 2y = 4 \implies x + 2y - 4 = 0$.
• Final Answer: x + 2y - 4 = 0
Review Exercise 8 Q4
Review Exercise 8: Form Reduction
: Reduce $5x - 2y + 1 = 0$ into slope-intercept form and two-intercept form.
Detailed Step-by-Step Solution:
• Step 1: $2y = 5x + 1 \implies y = \frac{5}{2}x + \frac{1}{2}$ (Slope-intercept form).
• Step 2: $5x - 2y = -1 \implies \frac{x}{-1/5} + \frac{y}{1/2} = 1$ (Two-intercept form).
• Final Answer: Slope-int: y = 5/2 x + 1/2; Two-int: x/(-1/5) + y/(1/2) = 1
• Step 2: $5x - 2y = -1 \implies \frac{x}{-1/5} + \frac{y}{1/2} = 1$ (Two-intercept form).
• Final Answer: Slope-int: y = 5/2 x + 1/2; Two-int: x/(-1/5) + y/(1/2) = 1
Review Exercise 8 Q5 (i)
Review Exercise 8: Intercept & Point-Slope Transformation
: Reduce $x - 3y = 3$ into intercept form and find $x$- and $y$-intercepts.
Detailed Step-by-Step Solution:
• Step 1: Divide by $3$: $\frac{x}{3} - \frac{3y}{3} = 1 \implies \frac{x}{3} + \frac{y}{-1} = 1$.
• Step 2: $x$-intercept $a = 3$, $y$-intercept $b = -1$.
• Final Answer: \frac{x}{3} + \frac{y}{-1} = 1; a = 3, b = -1
• Step 2: $x$-intercept $a = 3$, $y$-intercept $b = -1$.
• Final Answer: \frac{x}{3} + \frac{y}{-1} = 1; a = 3, b = -1
Review Exercise 8 Q5 (ii)
Review Exercise 8: Intercept & Point-Slope Transformation
: Find slope of $x - 3y = 3$ and transform the equation into point-slope form.
Detailed Step-by-Step Solution:
• Step 1: Slope $m = -\frac{A}{B} = -\frac{1}{-3} = \frac{1}{3}$.
• Step 2: Using point $(3, 0)$: $y - 0 = \frac{1}{3}(x - 3)$.
• Final Answer: m = 1/3; y - 0 = 1/3 (x - 3)
• Step 2: Using point $(3, 0)$: $y - 0 = \frac{1}{3}(x - 3)$.
• Final Answer: m = 1/3; y - 0 = 1/3 (x - 3)
Review Exercise 8 Q6
Review Exercise 8: Normal to Slope-Intercept Form
: Normal form of equation of line is $x \cos 150^\circ + y \sin 150^\circ = 10$. Transform into slope-intercept form and find slope and y-intercept.
Detailed Step-by-Step Solution:
• Step 1: $\cos 150^\circ = -\frac{\sqrt{3}}{2}, \sin 150^\circ = \frac{1}{2}$.
• Step 2: $-\frac{\sqrt{3}}{2}x + \frac{1}{2}y = 10 \implies -\sqrt{3}x + y = 20 \implies y = \sqrt{3}x + 20$.
• Step 3: Slope $m = \sqrt{3}$, $y$-intercept $c = 20$.
• Final Answer: y = \sqrt{3}x + 20; m = \sqrt{3}, c = 20
• Step 2: $-\frac{\sqrt{3}}{2}x + \frac{1}{2}y = 10 \implies -\sqrt{3}x + y = 20 \implies y = \sqrt{3}x + 20$.
• Step 3: Slope $m = \sqrt{3}$, $y$-intercept $c = 20$.
• Final Answer: y = \sqrt{3}x + 20; m = \sqrt{3}, c = 20
Review Exercise 8 Q7 (i)
Review Exercise 8: Triangle Geometry Graph Analysis
: From the textbook graph, find the coordinates of the vertices of $\Delta ABC$.
Detailed Step-by-Step Solution:
• Step 1: Reading from coordinate grid: $A(-4, 0), B(6, 2), C(0, 6)$.
• Final Answer: A(-4, 0), B(6, 2), C(0, 6)
• Final Answer: A(-4, 0), B(6, 2), C(0, 6)
Review Exercise 8 Q7 (ii)
Review Exercise 8: Triangle Geometry Graph Analysis
: Calculate slopes of sides of $\Delta ABC$ with $A(-4, 0), B(6, 2), C(0, 6)$.
Detailed Step-by-Step Solution:
• Step 1: $m_{AB} = \frac{2 - 0}{6 - (-4)} = \frac{2}{10} = \frac{1}{5}$.
• Step 2: $m_{BC} = \frac{6 - 2}{0 - 6} = \frac{4}{-6} = -\frac{2}{3}$.
• Step 3: $m_{CA} = \frac{0 - 6}{-4 - 0} = \frac{-6}{-4} = \frac{3}{2}$.
• Final Answer: m_AB = 1/5, m_BC = -2/3, m_CA = 3/2
• Step 2: $m_{BC} = \frac{6 - 2}{0 - 6} = \frac{4}{-6} = -\frac{2}{3}$.
• Step 3: $m_{CA} = \frac{0 - 6}{-4 - 0} = \frac{-6}{-4} = \frac{3}{2}$.
• Final Answer: m_AB = 1/5, m_BC = -2/3, m_CA = 3/2
Review Exercise 8 Q7 (iii)
Review Exercise 8: Triangle Geometry Graph Analysis
: Find interior angles of $\Delta ABC$ with $A(-4, 0), B(6, 2), C(0, 6)$.
Detailed Step-by-Step Solution:
• Step 1: Notice $m_{BC} \cdot m_{CA} = \left(-\frac{2}{3}\right)\left(\frac{3}{2}\right) = -1 \implies \angle C = 90^\circ$.
• Step 2: $\tan A = \left|\frac{3/2 - 1/5}{1 + (3/2)(1/5)}\right| = \frac{13/10}{13/10} = 1 \implies \angle A = 45^\circ$.
• Step 3: $\angle B = 180^\circ - (90^\circ + 45^\circ) = 45^\circ$.
• Final Answer: \angle A = 45^\circ, \angle B = 45^\circ, \angle C = 90^\circ
• Step 2: $\tan A = \left|\frac{3/2 - 1/5}{1 + (3/2)(1/5)}\right| = \frac{13/10}{13/10} = 1 \implies \angle A = 45^\circ$.
• Step 3: $\angle B = 180^\circ - (90^\circ + 45^\circ) = 45^\circ$.
• Final Answer: \angle A = 45^\circ, \angle B = 45^\circ, \angle C = 90^\circ
Review Exercise 8 Q8 (i)
Review Exercise 8: Line Properties & Equations
: Two points $P(4, -1)$ and $Q(8, 3)$ lie on a line. Find coordinates of mid-point $M$ of $PQ$.
Detailed Step-by-Step Solution:
• Step 1: $M = \left(\frac{4 + 8}{2}, \frac{-1 + 3}{2}\right) = (6, 1)$.
• Final Answer: M(6, 1)
• Final Answer: M(6, 1)
Review Exercise 8 Q8 (ii)
Review Exercise 8: Line Properties & Equations
: Two points $P(4, -1)$ and $Q(8, 3)$ lie on a line with midpoint $M(6, 1)$. Find slope of $PM$.
Detailed Step-by-Step Solution:
• Step 1: Slope of $PM = \frac{1 - (-1)}{6 - 4} = \frac{2}{2} = 1$.
• Final Answer: m = 1
• Final Answer: m = 1
Review Exercise 8 Q8 (iii)
Review Exercise 8: Line Properties & Equations
: Find equation of line parallel to $PQ$ through $(-2, 2)$ where $P(4, -1)$ and $Q(8, 3)$.
Detailed Step-by-Step Solution:
• Step 1: $m = 1$.
• Step 2: $y - 2 = 1(x - (-2)) \implies y - 2 = x + 2 \implies x - y + 4 = 0$.
• Final Answer: x - y + 4 = 0
• Step 2: $y - 2 = 1(x - (-2)) \implies y - 2 = x + 2 \implies x - y + 4 = 0$.
• Final Answer: x - y + 4 = 0
Review Exercise 8 Q8 (iv)
Review Exercise 8: Line Properties & Equations
: Find equation of line perpendicular to $PQ$ through $(-2, 2)$ where $P(4, -1)$ and $Q(8, 3)$.
Detailed Step-by-Step Solution:
• Step 1: $m_{\perp} = -1$.
• Step 2: $y - 2 = -1(x + 2) \implies y - 2 = -x - 2 \implies x + y = 0$.
• Final Answer: x + y = 0
• Step 2: $y - 2 = -1(x + 2) \implies y - 2 = -x - 2 \implies x + y = 0$.
• Final Answer: x + y = 0
Review Exercise 8 Q9 (i)
Review Exercise 8: Line Parameters Identification
: Points $A(2, -2)$ and $B(4, 6)$ lie on a line. Find length of $AB$.
Detailed Step-by-Step Solution:
• Step 1: $d = \sqrt{(4-2)^2 + (6 - (-2))^2} = \sqrt{2^2 + 8^2} = \sqrt{4 + 64} = \sqrt{68} = 2\sqrt{17}$.
• Final Answer: 2\sqrt{17} \approx 8.246
• Final Answer: 2\sqrt{17} \approx 8.246
Review Exercise 8 Q9 (ii)
Review Exercise 8: Line Parameters Identification
: Points $A(2, -2)$ and $B(4, 6)$ lie on a line. Find slope of $BA$.
Detailed Step-by-Step Solution:
• Step 1: $m = \frac{-2 - 6}{2 - 4} = \frac{-8}{-2} = 4$.
• Final Answer: m = 4
• Final Answer: m = 4
Review Exercise 8 Q9 (iii)
Review Exercise 8: Line Parameters Identification
: Find values of $a$ and $b$ when line $AB$ passing through $A(2, -2)$ and $B(4, 6)$ is written as $ax + by - 10 = 0$.
Detailed Step-by-Step Solution:
• Step 1: Line equation: $y - 6 = 4(x - 4) \implies 4x - y - 10 = 0$.
• Step 2: Comparing with $ax + by - 10 = 0 \implies a = 4, b = -1$.
• Final Answer: a = 4, b = -1
• Step 2: Comparing with $ax + by - 10 = 0 \implies a = 4, b = -1$.
• Final Answer: a = 4, b = -1
Review Exercise 8 Q9 (iv)
Review Exercise 8: Line Parameters Identification
: Find equation of line parallel to $AB$ passing through $(0, 3)$ where $A(2, -2)$ and $B(4, 6)$.
Detailed Step-by-Step Solution:
• Step 1: $m = 4$.
• Step 2: $y = 4x + 3 \implies 4x - y + 3 = 0$.
• Final Answer: 4x - y + 3 = 0
• Step 2: $y = 4x + 3 \implies 4x - y + 3 = 0$.
• Final Answer: 4x - y + 3 = 0
Review Exercise 8 Q10
Review Exercise 8: Midpoint & Slope Application
: Find equation of line passing through mid-point of $(4, 4)$ and $(8, 0)$ parallel to line having slope $\frac{2}{5}$.
Detailed Step-by-Step Solution:
• Step 1: Midpoint $M = \left(\frac{4+8}{2}, \frac{4+0}{2}\right) = (6, 2)$.
• Step 2: Slope $m = \frac{2}{5}$.
• Step 3: $y - 2 = \frac{2}{5}(x - 6) \implies 5y - 10 = 2x - 12 \implies 2x - 5y - 2 = 0$.
• Final Answer: 2x - 5y - 2 = 0
• Step 2: Slope $m = \frac{2}{5}$.
• Step 3: $y - 2 = \frac{2}{5}(x - 6) \implies 5y - 10 = 2x - 12 \implies 2x - 5y - 2 = 0$.
• Final Answer: 2x - 5y - 2 = 0
Review Exercise 8 Q11 (i)
Review Exercise 8: Graph Lines Analysis
: From the textbook graph, find the coordinates of end points of $OC$ and $AB$.
Detailed Step-by-Step Solution:
• Step 1: Reading from graph: $O(0, 0), C(6, 6), A(-4, 3), B(6, 1)$.
• Final Answer: O(0, 0), C(6, 6); A(-4, 3), B(6, 1)
• Final Answer: O(0, 0), C(6, 6); A(-4, 3), B(6, 1)
Review Exercise 8 Q11 (ii)
Review Exercise 8: Graph Lines Analysis
: From graph lines $OC$ [$O(0, 0), C(6, 6)$] and $AB$ [$A(-4, 3), B(6, 1)$], calculate slopes of both lines.
Detailed Step-by-Step Solution:
• Step 1: $m_{OC} = \frac{6 - 0}{6 - 0} = 1$.
• Step 2: $m_{AB} = \frac{1 - 3}{6 - (-4)} = \frac{-2}{10} = -\frac{1}{5}$.
• Final Answer: m_OC = 1, m_AB = -1/5
• Step 2: $m_{AB} = \frac{1 - 3}{6 - (-4)} = \frac{-2}{10} = -\frac{1}{5}$.
• Final Answer: m_OC = 1, m_AB = -1/5
Review Exercise 8 Q11 (iii)
Review Exercise 8: Graph Lines Analysis
: Find equations of both lines $OC$ and $AB$.
Detailed Step-by-Step Solution:
• Step 1: Line $OC$: $y = 1x \implies x - y = 0$.
• Step 2: Line $AB$: $y - 1 = -\frac{1}{5}(x - 6) \implies 5y - 5 = -x + 6 \implies x + 5y - 11 = 0$.
• Final Answer: Line OC: x - y = 0; Line AB: x + 5y - 11 = 0
• Step 2: Line $AB$: $y - 1 = -\frac{1}{5}(x - 6) \implies 5y - 5 = -x + 6 \implies x + 5y - 11 = 0$.
• Final Answer: Line OC: x - y = 0; Line AB: x + 5y - 11 = 0
Review Exercise 8 Q11 (iv)
Review Exercise 8: Graph Lines Analysis
: Find coordinates of the point of intersection of lines $OC$ and $AB$.
Detailed Step-by-Step Solution:
• Step 1: Line $OC$ is $y = x$. Substitute into $x + 5y - 11 = 0 \implies x + 5x = 11 \implies 6x = 11 \implies x = \frac{11}{6}$.
• Step 2: $y = \frac{11}{6}$. Intersection is $\left(\frac{11}{6}, \frac{11}{6}\right)$.
• Final Answer: (11/6, 11/6)
• Step 2: $y = \frac{11}{6}$. Intersection is $\left(\frac{11}{6}, \frac{11}{6}\right)$.
• Final Answer: (11/6, 11/6)
Review Exercise 8 Q12 (i)
Review Exercise 8: Segment Analysis from Equation
: Locate two points on the line $x - 2y = 2$ and find the slope of segment $l$ connecting them.
Detailed Step-by-Step Solution:
• Step 1: Let $y = 0 \implies x = 2 \implies (2, 0)$. Let $y = 1 \implies x = 4 \implies (4, 1)$.
• Step 2: Slope $m = \frac{1 - 0}{4 - 2} = \frac{1}{2}$.
• Final Answer: Points (2, 0) and (4, 1); Slope = 1/2
• Step 2: Slope $m = \frac{1 - 0}{4 - 2} = \frac{1}{2}$.
• Final Answer: Points (2, 0) and (4, 1); Slope = 1/2
Review Exercise 8 Q12 (ii)
Review Exercise 8: Segment Analysis from Equation
: For segment $l$ with slope $\frac{1}{2}$, find slope of segment $p$ perpendicular to $l$.
Detailed Step-by-Step Solution:
• Step 1: $m_p = -\frac{1}{m_l} = -\frac{1}{1/2} = -2$.
• Final Answer: m_p = -2
• Final Answer: m_p = -2
Review Exercise 8 Q12 (iii)
Review Exercise 8: Segment Analysis from Equation
: Find mid-point of the segment connecting $(2, 0)$ and $(4, 1)$.
Detailed Step-by-Step Solution:
• Step 1: $M = \left(\frac{2 + 4}{2}, \frac{0 + 1}{2}\right) = \left(3, \frac{1}{2}\right)$.
• Final Answer: M(3, 1/2)
• Final Answer: M(3, 1/2)
Review Exercise 8 Q12 (iv)
Review Exercise 8: Segment Analysis from Equation
: Find equation of line passing through mid-point $\left(3, \frac{1}{2}\right)$ with slope of $p$ ($m_p = -2$).
Detailed Step-by-Step Solution:
• Step 1: $y - \frac{1}{2} = -2(x - 3) \implies y - \frac{1}{2} = -2x + 6$.
• Step 2: Multiply by 2: $2y - 1 = -4x + 12 \implies 4x + 2y - 13 = 0$.
• Final Answer: 4x + 2y - 13 = 0
• Step 2: Multiply by 2: $2y - 1 = -4x + 12 \implies 4x + 2y - 13 = 0$.
• Final Answer: 4x + 2y - 13 = 0
Extra Exercise • Step-by-Step Complete Solutions
Extra Exercise Q1
Extra Exercise: Inclination Properties
: The inclination $\theta$ of any straight line in the Cartesian plane always lies in the interval:
Detailed Step-by-Step Solution:
• Step 1: Inclination $\theta$ is measured anti-clockwise from the positive x-axis and satisfies $0^\circ \le \theta < 180^\circ$.
• Final Answer: (b) 0^\circ \le \theta < 180^\circ
• Final Answer: (b) 0^\circ \le \theta < 180^\circ
Extra Exercise Q2
Extra Exercise: Horizontal Line Inclination
: What is the inclination of any line parallel to the x-axis?
Detailed Step-by-Step Solution:
• Step 1: A horizontal line is parallel to the x-axis, making an angle of $0^\circ$.
• Final Answer: (a) 0^\circ
• Final Answer: (a) 0^\circ
Extra Exercise Q3
Extra Exercise: Perpendicularity Test
: If two lines with non-zero slopes $m_1$ and $m_2$ are perpendicular, which relationship holds?
Detailed Step-by-Step Solution:
• Step 1: Two non-vertical lines are perpendicular if and only if $m_1 m_2 = -1$.
• Final Answer: (c) m_1 \cdot m_2 = -1
• Final Answer: (c) m_1 \cdot m_2 = -1
Extra Exercise Q4
Extra Exercise: Slope from General Linear Form
: What is the slope of the general linear equation $Ax + By + C = 0$ ($B \ne 0$)?
Detailed Step-by-Step Solution:
• Step 1: $By = -Ax - C \implies y = -\frac{A}{B}x - \frac{C}{B}$. Thus slope $m = -\frac{A}{B}$.
• Final Answer: (a) -A/B
• Final Answer: (a) -A/B
Extra Exercise Q5
Extra Exercise: Y-Intercept from General Form
: What is the y-intercept of the line $Ax + By + C = 0$ ($B \ne 0$)?
Detailed Step-by-Step Solution:
• Step 1: Set $x = 0 \implies By + C = 0 \implies y = -\frac{C}{B}$.
• Final Answer: (b) -C/B
• Final Answer: (b) -C/B
Extra Exercise Q6
Extra Exercise: X-Intercept from General Form
: What is the x-intercept of the line $Ax + By + C = 0$ ($A \ne 0$)?
Detailed Step-by-Step Solution:
• Step 1: Set $y = 0 \implies Ax + C = 0 \implies x = -\frac{C}{A}$.
• Final Answer: (a) -C/A
• Final Answer: (a) -C/A
Extra Exercise Q7
Extra Exercise: Two-Intercept Form
: The equation $\frac{x}{a} + \frac{y}{b} = 1$ is known as:
Detailed Step-by-Step Solution:
• Step 1: This is the standard Two-Intercept Form where $a$ and $b$ are the x- and y-intercepts.
• Final Answer: (c) Two-intercept form
• Final Answer: (c) Two-intercept form
Extra Exercise Q8
Extra Exercise: Area with Axes
: The area of the triangle formed by the line $\frac{x}{a} + \frac{y}{b} = 1$ with the coordinate axes is:
Detailed Step-by-Step Solution:
• Step 1: The vertices are $(0, 0), (a, 0), (0, b)$. Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}|ab|$.
• Final Answer: (b) 1/2 |ab|
• Final Answer: (b) 1/2 |ab|
Extra Exercise Q9
Extra Exercise: Normal Form Parameter
: In the normal form $x \cos \alpha + y \sin \alpha = p$, the parameter $p$ represents:
Detailed Step-by-Step Solution:
• Step 1: $p$ is the length of the normal from origin $(0, 0)$ to the line ($p \ge 0$).
• Final Answer: (b) Perpendicular distance from origin
• Final Answer: (b) Perpendicular distance from origin
Extra Exercise Q10
Extra Exercise: Normal Distance Value
: The perpendicular distance $p$ from the origin to $3x + 4y - 20 = 0$ is:
Detailed Step-by-Step Solution:
• Step 1: $p = \frac{|-20|}{\sqrt{3^2 + 4^2}} = \frac{20}{5} = 4$.
• Final Answer: (b) 4
• Final Answer: (b) 4
Extra Exercise Q11
Extra Exercise: Symmetric Form Components
: In symmetric form $\frac{x - x_1}{\cos \alpha} = \frac{y - y_1}{\sin \alpha} = r$, what does $r$ represent?
Detailed Step-by-Step Solution:
• Step 1: $r$ represents the algebraic directed distance from $(x_1, y_1)$ to any point $(x, y)$ along the line.
• Final Answer: (b) Directed distance from (x_1, y_1)
• Final Answer: (b) Directed distance from (x_1, y_1)
Extra Exercise Q12
Extra Exercise: Parallel Lines Slope Condition
: Two non-vertical lines $l_1$ and $l_2$ are parallel if and only if:
Detailed Step-by-Step Solution:
• Step 1: Parallel lines share identical inclination and slope: $m_1 = m_2$.
• Final Answer: (b) m_1 = m_2
• Final Answer: (b) m_1 = m_2
Extra Exercise Q13
Extra Exercise: Angle Between Lines Formula
: The tangent of angle $\theta$ from line $l_1$ to line $l_2$ is given by:
Detailed Step-by-Step Solution:
• Step 1: Standard formula: $\tan \theta = \frac{m_2 - m_1}{1 + m_1 m_2}$.
• Final Answer: (a) (m_2 - m_1)/(1 + m_1 m_2)
• Final Answer: (a) (m_2 - m_1)/(1 + m_1 m_2)
Extra Exercise Q14
Extra Exercise: Family of Lines
: The equation $L_1 + k L_2 = 0$ represents:
Detailed Step-by-Step Solution:
• Step 1: $L_1 + k L_2 = 0$ generates every straight line passing through the common intersection of $L_1$ and $L_2$.
• Final Answer: (b) A family of lines through intersection of L_1 and L_2
• Final Answer: (b) A family of lines through intersection of L_1 and L_2
Extra Exercise Q15
Extra Exercise: Collinearity Condition
: Three distinct points $A, B, C$ are collinear if and only if:
Detailed Step-by-Step Solution:
• Step 1: Common slope through a shared point proves all three points lie on the same straight line.
• Final Answer: (a) Slope(AB) = Slope(BC)
• Final Answer: (a) Slope(AB) = Slope(BC)
Extra Exercise Q16
Extra Exercise: Origin Line Constant
: Any straight line passing through the origin $(0, 0)$ has its constant term $C$ equal to:
Detailed Step-by-Step Solution:
• Step 1: Substituting $(0, 0)$ into $Ax + By + C = 0 \implies C = 0$.
• Final Answer: (c) 0
• Final Answer: (c) 0
Extra Exercise Q17
Extra Exercise: Line through Midpoint
: A line segment connects $(2, 4)$ and $(6, 8)$. What is the midpoint?
Detailed Step-by-Step Solution:
• Step 1: $M = \left(\frac{2+6}{2}, \frac{4+8}{2}\right) = (4, 6)$.
• Final Answer: (a) (4, 6)
• Final Answer: (a) (4, 6)
Extra Exercise Q18
Extra Exercise: Obtuse Angle Slope Sign
: If the inclination of a straight line is obtuse ($90^\circ < \theta < 180^\circ$), its slope is:
Detailed Step-by-Step Solution:
• Step 1: In Quadrant II, $\tan \theta < 0$, so slope is negative.
• Final Answer: (b) Negative
• Final Answer: (b) Negative
Extra Exercise Q19
Extra Exercise: Acute Angle Slope Sign
: If the inclination of a straight line is acute ($0^\circ < \theta < 90^\circ$), its slope is:
Detailed Step-by-Step Solution:
• Step 1: In Quadrant I, $\tan \theta > 0$, so slope is positive.
• Final Answer: (a) Positive
• Final Answer: (a) Positive
Extra Exercise Q20
Extra Exercise: Perpendicular Lines Identification
: Which pair of lines are perpendicular to each other?
Detailed Step-by-Step Solution:
• Step 1: $m_1 = 2, m_2 = -\frac{1}{2} \implies m_1 m_2 = 2\left(-\frac{1}{2}\right) = -1$.
• Final Answer: (b) y = 2x + 4 and y = -1/2 x + 7
• Final Answer: (b) y = 2x + 4 and y = -1/2 x + 7
Extra Exercise Q21
Extra Exercise: Intersection with Y-Axis
: The line $3x - 4y + 12 = 0$ crosses the y-axis at:
Detailed Step-by-Step Solution:
• Step 1: Set $x = 0 \implies -4y + 12 = 0 \implies y = 3$. Point is $(0, 3)$.
• Final Answer: (a) (0, 3)
• Final Answer: (a) (0, 3)
Extra Exercise Q22
Extra Exercise: Intersection with X-Axis
: The line $3x - 4y + 12 = 0$ crosses the x-axis at:
Detailed Step-by-Step Solution:
• Step 1: Set $y = 0 \implies 3x + 12 = 0 \implies x = -4$. Point is $(-4, 0)$.
• Final Answer: (a) (-4, 0)
• Final Answer: (a) (-4, 0)
Extra Exercise Q23
Extra Exercise: Parallel Line Coefficient Match
: Any line parallel to $5x - 7y + 9 = 0$ must have the form:
Detailed Step-by-Step Solution:
• Step 1: Parallel lines share the same ratio of $x$ and $y$ coefficients: $5x - 7y + k = 0$.
• Final Answer: (a) 5x - 7y + k = 0
• Final Answer: (a) 5x - 7y + k = 0
Extra Exercise Q24
Extra Exercise: Perpendicular Line Coefficient Match
: Any line perpendicular to $5x - 7y + 9 = 0$ must have the form:
Detailed Step-by-Step Solution:
• Step 1: Perpendicular line swaps coefficients and negates one sign: $7x + 5y + k = 0$.
• Final Answer: (a) 7x + 5y + k = 0
• Final Answer: (a) 7x + 5y + k = 0
Extra Exercise Q25
Extra Exercise: Real World Slope Meaning
: In a real-world cost model $C = 25n + 500$, the number $25$ represents:
Detailed Step-by-Step Solution:
• Step 1: $25$ is the rate of change (slope), representing the additional cost per unit produced.
• Final Answer: (b) Marginal cost per unit produced (Slope)
• Final Answer: (b) Marginal cost per unit produced (Slope)
🎯 7. Unit Synthesis Summary
Chapter 8 masterfully connects algebraic linear equations with geometric straight lines. A line is uniquely governed by its steepness ($m = \tan \theta$) and its position relative to the Cartesian origin. Mastery of the 6 standard forms ($y=mx+c$, $y-y_1=m(x-x_1)$, two-point, two-intercept, symmetric, and normal forms) empowers students to effortlessly analyze parallelism, perpendicularity, geometric angles, intersections, and dynamic real-world linear phenomena.
More Chapter Notes for Class 9 (FBISE)
Mathematics
Mathematics • Chapter 1
FBISE
Mastery Guide: Real Numbers — Classification, Number Line, Radicals & Laws of Exponents
Mathematics • Chapter 2
FBISE
Unit 02: Logarithms
Mathematics • Chapter 3
FBISE
Mastery Guide: Sets and Relations — Set Operations, Venn Diagrams, Survey Inclusion-Exclusion, Cartesian Products & Binary Relations
Mathematics • Chapter 4
FBISE
Mastery Guide: Factorization, HCF, LCM & Algebraic Fractions
Mathematics • Chapter 5
FBISE
Mastery Guide: Linear Equations, Radicals, Absolute Values & Inequalities
Mathematics • Chapter 6
FBISE
Mastery Guide: Trigonometry & Bearing — Angle Systems, Circle Sectors, Unit Circle Ratios, Fundamental Identities, Real-World Heights & Distances, and 3-Digit True Bearings
Mathematics • Chapter 7
FBISE
Mastery Guide: Coordinate Geometry — 1D/2D Distance Formula, Collinearity, Polygon Classifications, Mid-Point Formula & Midpoint Theorem
Mathematics • Chapter 9
FBISE
Mastery Guide: Geometry and Polygons - Demonstrative Geometry, Similarity of Figures, Regular Polygons & Scaling
Mathematics • Chapter 10
FBISE
Mastery Guide: Practical Geometry - Triangle Constructions, Ambiguous Case, Angle Bisectors, Altitudes, Perp Bisectors & Centers
Mathematics • Chapter 11
FBISE
Mastery Guide: Basic Statistics - Frequency Distributions, Histograms, Central Tendencies & Probability
Self-Assessment Practice
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