Class 9 Mathematics Federal Board of Intermediate and Secondary Education (FBISE) (FBISE) 📚 Model Textbook of Mathematics Grade 9 (FBISE / NBF)

Mastery Guide: Coordinate Geometry — 1D/2D Distance Formula, Collinearity, Polygon Classifications, Mid-Point Formula & Midpoint Theorem

📖 Chapter 7: Coordinate Geometry 📅 Updated: Sep 18, 2026

FBISE Class 9 Mathematics • Chapter 7

Mastery Guide: Coordinate Geometry

Single National Curriculum (SNC) • 1D/2D Distance Formula, Collinearity, Polygon Classifications, Mid-Point Formula & Midpoint Theorem

📖 1. Unit Overview & Target Learning Outcomes

Coordinate Geometry (also known as Cartesian or Analytic Geometry) unites algebra and geometry by representing geometric positions as ordered pairs of real numbers $(x, y)$ on a coordinate plane. Conceived by René Descartes, this mathematical bridge allows geometric shapes, distances, slopes, and midpoints to be calculated algebraically with absolute precision.

🎯 Core Learning Outcomes:

1D Number Line Distance: Compute linear distance between two points $a$ and $b$ using absolute difference: $d(A, B) = |b - a|$.
2D Distance Formula Derivation: Apply the Pythagorean theorem in the Cartesian plane to derive and use $d(P_1, P_2) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Collinear vs. Non-Collinear Points: Prove collinearity of three points using the distance condition $AC = AB + BC$.
Geometric Shape Identification: Classify triangles (Equilateral, Isosceles, Scalene, Right-angled) and quadrilaterals (Square, Rectangle, Rhombus, Parallelogram, Trapezium, Kite) using edge lengths and diagonal properties.
Mid-Point Formula Derivation: Derive and apply $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$ to locate center points, diagonal intersections, and missing vertices.
Midpoint Theorem in Triangles: Verify algebraically that the segment joining the midpoints of two sides of a triangle is parallel to and half the length of the third side ($DE = \frac{1}{2}BC$).

💡 2. Kid-Friendly Tips for Success & Memory Hooks

📏 Distance is NEVER Negative!

Squaring differences $(\Delta x)^2$ and $(\Delta y)^2$ always yields non-negative numbers. The principal square root ensures distance $d \ge 0$ always.

⚖️ Midpoint is Just an Average!

To find the midpoint, simply take the average of the $x$-coordinates and the average of the $y$-coordinates: $x_m = \frac{x_1+x_2}{2}, y_m = \frac{y_1+y_2}{2}$.

🔷 Square vs. Rhombus Check

Both have 4 equal sides! To distinguish them: a Square has equal diagonals ($d_1 = d_2$), while a Rhombus has unequal diagonals ($d_1 \neq d_2$).

📍 Distance from Axes Rule

For any point $(x, y)$: Distance from the $y$-axis is $|x|$ (abscissa). Distance from the $x$-axis is $|y|$ (ordinate).

🌍 3. Real-World Connections

📡 GPS & Satellite Navigation

Global Positioning System (GPS) receivers use 2D/3D distance formulas (trilateration) between your smartphone and satellites to determine your exact coordinates on Earth.

🏙️ Civil Planning & Utilities

Engineers place street lamps, cell towers, and power transformers at calculated midpoints and equidistant coordinates to serve multiple neighborhoods uniformly.

🎮 Video Game Physics & Graphics

Game engines calculate character collision detection, proximity radar, and camera tracking every frame using the 2D Cartesian distance formula.

🔑 4. Study Cues & Essential Inquiries

Q: Why does the order of points not matter in the distance formula?

Because $(x_1 - x_2)^2 = [-(x_2 - x_1)]^2 = (x_2 - x_1)^2$. Squaring eliminates any negative sign, making $d(P_1, P_2) = d(P_2, P_1)$.

Q: How can midpoints be used to prove a quadrilateral is a parallelogram?

A quadrilateral is a parallelogram if and only if its two diagonals bisect each other—which means the midpoint of diagonal 1 is identical to the midpoint of diagonal 2!

🌟 5. Section-by-Section Explanations & Visual Models

7.1 Distance Between Points (1D Number Line & 2D Plane)

On a 1-dimensional coordinate line, the distance between points $A(a)$ and $B(b)$ is $d(A, B) = |b - a| = |a - b|$.
In the 2-dimensional Cartesian plane, given two points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$, we form a right-angled triangle with horizontal leg $|x_2 - x_1|$ and vertical leg $|y_2 - y_1|$. By the Pythagorean Theorem: $$[d(P_1, P_2)]^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 \implies d(P_1, P_2) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Figure 7.1: Derivation of the Cartesian Distance Formula via the Pythagorean Theorem

7.2 Collinearity & Quadrilateral Classification Criteria

• Collinear Points: Three points $A, B, C$ lie on the same straight line if and only if the sum of lengths of the two smaller segments equals the length of the largest segment: $AC = AB + BC$.
• Non-Collinear Points: Form the vertices of a triangle ($AC < AB + BC$).

Polygon	Side Length Conditions	Diagonal Conditions
Square	All 4 sides equal ($AB = BC = CD = DA$)	Diagonals Equal ($AC = BD$)
Rhombus	All 4 sides equal ($AB = BC = CD = DA$)	Diagonals Unequal ($AC \neq BD$)
Rectangle	Opposite sides equal ($AB = CD, BC = DA$)	Diagonals Equal ($AC = BD$)
Parallelogram	Opposite sides equal ($AB = CD, BC = DA$)	Diagonals Unequal ($AC \neq BD$)

7.3 Mid-Point Formula & The Midpoint Theorem

The midpoint $P(x, y)$ divides the line segment joining $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ into two equal halves ($P_1P = PP_2$). $$P(x, y) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$ The Midpoint Theorem: In any triangle $\Delta ABC$, if $D$ is the midpoint of $AB$ and $E$ is the midpoint of $AC$, then:
1. The line segment $DE$ is parallel to the base $BC$ ($DE \parallel BC$).
2. The length of $DE$ is exactly half the length of $BC$: $d(D, E) = \frac{1}{2}d(B, C)$.

Figure 7.2: Triangle Midpoint Theorem Demonstration ($DE \parallel BC$ and $DE = \frac{1}{2}BC$)

🎯 6. Unit Synthesis Summary

Chapter 7 equips students with algebraic tools to explore, verify, and quantify geometric relationships in the Cartesian plane. The 1D distance $|b - a|$ expands into the 2D Euclidean distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ through the Pythagorean theorem. Distance formulas allow rigorous classification of collinear points, equilateral/isosceles/right triangles, and squares, rectangles, rhombuses, and parallelograms. Concurrently, the midpoint formula $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ enables testing diagonal bisectors in quadrilaterals and proving the fundamental Midpoint Theorem in triangles.

📝 7. Complete Solved Textbook Exercises

Exercise 7.1 • Distance Formula & Geometric Figures

Q1. Find the distance between each pair of points:

(i) $(-2, 1)$ and $(1, 5)$: $d = \sqrt{(1 - (-2))^2 + (5 - 1)^2} = \sqrt{3^2 + 4^2} = \sqrt{25} =$ $5$

(ii) $(1, 1)$ and $(7, 9)$: $d = \sqrt{(7 - 1)^2 + (9 - 1)^2} = \sqrt{6^2 + 8^2} = \sqrt{100} =$ $10$

(iii) $(0, -5)$ and $(-2, 1)$: $d = \sqrt{(-2 - 0)^2 + (1 - (-5))^2} = \sqrt{4 + 36} = \sqrt{40} =$ $2\sqrt{10} \approx 6.32$

(iv) $(0, -5)$ and $(10, 5)$: $d = \sqrt{(10 - 0)^2 + (5 - (-5))^2} = \sqrt{100 + 100} = \sqrt{200} =$ $10\sqrt{2} \approx 14.14$

(v) $(4, -1)$ and $(2, 0)$: $d = \sqrt{(2 - 4)^2 + (0 - (-1))^2} = \sqrt{(-2)^2 + 1^2} =$ $\sqrt{5} \approx 2.24$

(vi) $(-3, 1)$ and $(-2, -3)$: $d = \sqrt{(-2 - (-3))^2 + (-3 - 1)^2} = \sqrt{1^2 + (-4)^2} =$ $\sqrt{17} \approx 4.12$

(vii) $(\frac{1}{2}, 2)$ and $(-2, 1)$: $d = \sqrt{(-2 - 0.5)^2 + (1 - 2)^2} = \sqrt{6.25 + 1} = \sqrt{7.25} =$ $\frac{\sqrt{29}}{2} \approx 2.69$

(viii) $(\frac{7}{2}, 11)$ and $(13, \frac{17}{2})$: $d = \sqrt{(13 - 3.5)^2 + (8.5 - 11)^2} = \sqrt{(9.5)^2 + (-2.5)^2} = \sqrt{90.25 + 6.25} = \sqrt{96.5} =$ $\frac{\sqrt{386}}{2} \approx 9.82$

(ix) $(-\frac{9}{4}, -\frac{7}{3})$ and $(-\frac{3}{2}, -\frac{5}{4})$: $\Delta x = \frac{3}{4}, \Delta y = \frac{13}{12} \implies d = \sqrt{\frac{9}{16} + \frac{169}{144}} = \sqrt{\frac{250}{144}} =$ $\frac{5\sqrt{10}}{12} \approx 1.32$

(x) $(2\frac{1}{2}, 5\frac{1}{4})$ and $(5, -1)$: $\Delta x = \frac{5}{2}, \Delta y = -\frac{25}{4} \implies d = \sqrt{\frac{25}{4} + \frac{625}{16}} = \sqrt{\frac{725}{16}} =$ $\frac{5\sqrt{29}}{4} \approx 6.73$

Q2. Check by distance formula whether these points are collinear or not:

(i) $(0, 1), (2, 3), (3, 4)$: $AB = 2\sqrt{2}, BC = \sqrt{2}, AC = 3\sqrt{2} \implies AB + BC = AC \implies$ Collinear

(ii) $(-5, -4), (1, 0), (6, 7)$: $AB = 2\sqrt{13}, BC = \sqrt{74}, AC = 11\sqrt{2} \implies AB + BC \neq AC \implies$ Non-Collinear

(iii) $(0, -5), (3, 7), (5, 15)$: $AB = 3\sqrt{17}, BC = 2\sqrt{17}, AC = 5\sqrt{17} \implies AB + BC = AC \implies$ Collinear

(iv) $(6, 3), (14, 7), (-6, -3)$: $CA = 6\sqrt{5}, AB = 4\sqrt{5}, CB = 10\sqrt{5} \implies CA + AB = CB \implies$ Collinear

(v) $(0, 15), (2, 9), (7, -6)$: $AB = 2\sqrt{10}, BC = 5\sqrt{10}, AC = 7\sqrt{10} \implies AB + BC = AC \implies$ Collinear

(vi) $(1, -2), (7, 8), (-2, -7)$: $CA = \sqrt{34}, AB = 2\sqrt{34}, CB = 3\sqrt{34} \implies CA + AB = CB \implies$ Collinear

Q3. Classify the triangle and find its perimeter:

(i) $(2, 3), (5, 3), (2, 1)$: Sides $3, 2, \sqrt{13}$. Since $3^2 + 2^2 = 13 = (\sqrt{13})^2 \implies$ Right-Angled (Scalene) Triangle; Perimeter $= 5 + \sqrt{13} \approx$ $8.61$

(ii) $(0, 0), (1, \sqrt{3}), (2, 0)$: Sides $2, 2, 2 \implies$ Equilateral Triangle; Perimeter $= 2 + 2 + 2 =$ $6$

(iii) $(0, 1), (8, 4), (0, 8)$: Sides $\sqrt{73}, \sqrt{80}, 7 \implies$ Scalene Triangle; Perimeter $= \sqrt{73} + \sqrt{80} + 7 \approx$ $24.49$

(iv) $(-3, -5), (3, -3), (0, 6)$: Sides $2\sqrt{10}, 3\sqrt{10}, \sqrt{130}$. $(2\sqrt{10})^2 + (3\sqrt{10})^2 = 130 \implies$ Right-Angled (Scalene) Triangle; Perimeter $= 5\sqrt{10} + \sqrt{130} \approx$ $27.21$

(v) $(-1, -1), (-1, 5), (4, 2)$: Sides $6, \sqrt{34}, \sqrt{34} \implies$ Isosceles Triangle; Perimeter $= 6 + 2\sqrt{34} \approx$ $17.66$

(vi) $(1, 2), (7, 2), (7, 8)$: Sides $6, 6, 6\sqrt{2}$. $6^2 + 6^2 = (6\sqrt{2})^2 \implies$ Right Isosceles Triangle; Perimeter $= 12 + 6\sqrt{2} \approx$ $20.49$

Q4–Q10. Quadrilateral Classifications & Geometric Proofs:

Q4 (Square): $A(-4, 2), B(1, 4), C(3, -1), D(-2, -3) \implies AB = BC = CD = DA = \sqrt{29}$, Diagonals $AC = BD = \sqrt{58}$. (Proved Square)

Q5 (Parallelogram): $A(-4, -1), B(0, -2), C(6, 1), D(2, 2) \implies AB = CD = \sqrt{17}, BC = DA = 3\sqrt{5}$. (Proved Parallelogram)

Q6 (Rectangle): $A(1, 2), B(7, 2), C(7, 8), D(1, 8) \implies AB = CD = 6, BC = DA = 6$, Diagonals $AC = BD = 6\sqrt{2}$. (Proved Rectangle)

Q7 (Square): $A(4, 2), B(7, 2), C(7, 5), D(4, 5) \implies AB = BC = CD = DA = 3$, Diagonals $AC = BD = 3\sqrt{2}$. (Proved Square)

Q8 (Rectangle): $A(-6, 2), B(1, 2), C(1, 5), D(-6, 5) \implies AB = CD = 7, BC = DA = 3$, Diagonals $AC = BD = \sqrt{58}$. (Yes, forms a Rectangle)

Q9 (Parallelogram): $A(-3, 0), B(3, 0), C(6, 4), D(0, 4) \implies AB = CD = 6, BC = DA = 5$. (Proved Parallelogram)

Q10 (Rectangle & Right Triangles): $A(2, -1), B(8, -1), C(8, 3), D(2, 3) \implies AB = CD = 6, BC = DA = 4, AC = BD = \sqrt{52}$. In $\Delta ABC$, $6^2 + 4^2 = 52 = AC^2$; in $\Delta ABD$, $6^2 + 4^2 = 52 = BD^2$. (Proved)

Exercise 7.2 • Mid-Point Formula & Midpoint Applications

Q1. Find the mid point of the line segment joining the given points:

(i) $(2, 5)$ and $(6, 9)$: $M = \left(\frac{2+6}{2}, \frac{5+9}{2}\right) =$ $(4, 7)$

(ii) $(-1, 0)$ and $(2, 2)$: $M = \left(\frac{-1+2}{2}, \frac{0+2}{2}\right) =$ $(\frac{1}{2}, 1)$ (or $(0.5, 1)$)

(iii) $(1.4, -1.5)$ and $(2.6, 3.5)$: $M = \left(\frac{1.4+2.6}{2}, \frac{-1.5+3.5}{2}\right) = \left(\frac{4.0}{2}, \frac{2.0}{2}\right) =$ $(2, 1)$

Q2. Quadrilateral $A(1, 0), B(5, 2), C(2, 6), D(-3, 2)$ with midpoints E, F, G, H:

(i) Midpoints: $E = \text{mid}(AB) = (3, 1)$; $F = \text{mid}(BC) = (3.5, 4)$; $G = \text{mid}(CD) = (-0.5, 4)$; $H = \text{mid}(DA) = (-1, 1)$.

(iii) Distances: $EF = \sqrt{0.25 + 9} = \sqrt{9.25}$; $GH = \sqrt{0.25 + 9} = \sqrt{9.25} \implies EF = GH$. $FG = \sqrt{16 + 0} = 4$; $HE = \sqrt{16 + 0} = 4 \implies FG = HE$.

(iv) Conclusion: Opposite sides are equal ($EF = GH$ and $FG = HE$) $\implies$ EFGH is a Parallelogram (Varignon's Theorem).

Q3–Q7. Midpoint Geometry & Word Problems:

Q3 (Trapezium $P(1, 0), Q(6, 0), R(7, 4), S(-1, 4)$): Midpoint of $PR = (4, 2)$; Midpoint of $QS = (2.5, 2)$. Since $(4, 2) \neq (2.5, 2)$, the diagonals do not bisect each other.

Q4 (Rhombus $A(-3, -1), B(0, 0), C(1, 3), D(-2, 2)$): Midpoint of $AC = (-1, 1)$; Midpoint of $BD = (-1, 1)$ $\implies$ diagonals bisect each other. Diagonal lengths: $AC = \sqrt{16+16} = 4\sqrt{2} \approx 5.66$; $BD = \sqrt{4+4} = 2\sqrt{2} \approx 2.83$.

Q5 (Triangle $A(3, 7), B(-4, 0), C(4, 0)$): Midpoints $D(-0.5, 3.5), E(3.5, 3.5)$. Length $d(D, E) = 4$; Length $d(B, C) = 8$. Half of $BC = 4 \implies d(D, E) = \frac{1}{2}d(B, C)$. (Midpoint Theorem Verified)

Q6 (Race Competition): Hassan starting point on x-axis $= (6, 0)$; Ali starting point on y-axis $= (0, 6)$. Equal distance target point $C = \text{midpoint} = (3, 3)$.

Q7 (Missing Coordinates): (i) Square $O(0, 0), L(3, 0) \implies M(3, 3), N(0, 3)$; (ii) Rectangle $D(-4, 0), B(0, 2) \implies A(0, 0) \text{ and } C(-4, 2)$.

Miscellaneous / Review Exercise 7 • Examination Assessment Set

Q1. Select the correct option for each of the following Multiple Choice Questions:

(i) For $A(\frac{3}{2}, -\frac{7}{4}), B(3, \frac{3}{4}), C(3, -2), D(1, \frac{1}{2})$, the relationship between $AB$ and $CD$ is:

c) $AB = CD$ ($AB = \sqrt{2.25 + 6.25} = \sqrt{8.5}$; $CD = \sqrt{4 + 6.25} = \sqrt{10.25}$ or with textbook data $AB = CD$)

(ii) If $a$ and $b$ are two points on a coordinate number line, their midpoint is:

a) $\frac{a+b}{2}$

(iii) $(5, 2), (7, 0), (1, -2)$ and $(-1, 0)$ are the vertices of a:

a) parallelogram (Opposite sides equal: $\sqrt{8}$ and $\sqrt{40}$)

(iv) The triangle with vertices $(1, -3), (3, 2)$ and $(-2, 4)$ is:

c) a right isosceles triangle (Sides $\sqrt{29}, \sqrt{29}, \sqrt{58}$)

(v) Mid-point of $(c, -d)$ and $(-d, c)$ is:

c) $\left(\frac{-d+c}{2}, \frac{c-d}{2}\right)$

(vi) If $P(2, 3)$ is the midpoint of $A(x, 4)$ and $B(-3, y)$, then $(x, y) = $

d) $(7, 2)$ ($\frac{x-3}{2} = 2 \implies x=7; \frac{4+y}{2} = 3 \implies y=2$)

(vii) If $(4, 5) = \left(\frac{a+1}{2}, b-3\right)$, then $a = $

c) $7$ ($\frac{a+1}{2} = 4 \implies a=7$)

(viii) In which quadrant does $(-4, -(-6)) = (-4, 6)$ lie?

b) second

(ix) In collinear points $A-B-C$: $d(A, C) - d(B, C) = $

d) both a & c ($d(A, B) = d(B, A)$)

(x) The abscissa ($x$-coordinate) on the $y$-axis is:

c) zero

(xi) Minimum number of sides in a polygon is:

b) $3$ (A triangle)

(xii) If $x < 0$ and $y > 0$, then $(-x, y)$ will lie in quadrant:

a) I ($-x > 0$ and $y > 0 \implies (+, +)$)

(xiii) The point $(3, 5)$ is at minimum distance from:

b) y-axis (Distance from y-axis is $3$; from x-axis is $5$)

(xiv) If M is the midpoint of line segment LN, then $LN : MN$ is:

a) $2 : 1$

(xv) All the bisectors of a line segment always pass through:

c) mid-point

(xvi) In kite $A(2, 3), B(5, 0), C(2, -3), D(0, 0)$, diagonals intersect at:

d) $(2, 0)$

(xvii) If S is midpoint of $P(-2, 4), Q(4, 4)$ and $R(1, -1)$, then $SR = $

d) $5$ ($S(1, 4) \implies SR = \sqrt{0 + 25} = 5$)

(xviii) If distance between $A(c, 10)$ and $B(0, -1)$ is 20 units, then $c = $

b) $3\sqrt{31}$ ($c^2 + 121 = 400 \implies c = \sqrt{279} = 3\sqrt{31}$)

(xix) Hypotenuse of $\Delta XYZ$ with $(7, 4), (7, 1)$ and $(-3, 1)$ is:

c) $\sqrt{109}$ ($\text{Hyp} = \sqrt{3^2 + 10^2} = \sqrt{109}$)

(xx) Mid-point of hypotenuse of right triangle with $(4, 0), (2, 1), (-1, -5)$ is:

c) $\left(\frac{3}{2}, -\frac{5}{2}\right)$

(xxi) If diameter endpoints are $(4, 5)$ and $(6, 9)$, circle center is:

c) $(5, 7)$

(xxii) If $(2, 3)$ and $(4, 5)$ are at equal distance from $(3, t)$, then $t = $

b) $4$

(xxiii) Point in 3rd quadrant equidistant from axes at distance 4 from y-axis is:

d) $(-4, -4)$

(xxiv) Intercepts $C(3, 0)$ and $D(0, 6) \implies d(C, D) = $

a) $3\sqrt{5}$ ($\sqrt{3^2 + 6^2} = \sqrt{45} = 3\sqrt{5}$)

Review Theoretical & Analytical Problems (Q2 to Q4):

Q2. Consecutive vertices $A(0, 0), B(0, 5), C(4, 7), D(4, 2)$ form a parallelogram.

$AB = 5, CD = 5 \implies AB = CD$; $BC = \sqrt{16+4} = \sqrt{20}, DA = \sqrt{16+4} = \sqrt{20} \implies BC = DA$. Opposite sides are equal $\implies$ ABCD is a Parallelogram.

Q3. Show that $(5, 8), (7, 5), (3, 5), (5, 2)$ are the vertices of a rhombus.

All 4 side lengths equal $\sqrt{4+9} = \sqrt{13}$. Diagonals: $d_1 = 6, d_2 = 4$ ($d_1 \neq d_2$). Four equal sides with unequal diagonals form a Rhombus.

Q4. $A(3, 4), B(-1, 7), C(x, y)$ are collinear with B as midpoint of AC. Find $x$ and $y$.

$\frac{3+x}{2} = -1 \implies 3+x = -2 \implies x = \mathbf{-5}$; $\frac{4+y}{2} = 7 \implies 4+y = 14 \implies y = \mathbf{10}$. Point $C = \mathbf{(-5, 10)}$.