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Class 9 Mathematics — Chapter 7: Coordinate Geometry

Punjab Board | 2026

Exercise 7.1

Q1. Find the distance between the following pairs of points.

Distance Formula

\[ d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \]

(i) A(2, 3), B(6, 6)

Step 1: Identify the coordinates.

\[ x_1=2,\quad y_1=3,\quad x_2=6,\quad y_2=6 \]

Step 2: Apply the distance formula.

\[ d=\sqrt{(6-2)^2+(6-3)^2} \]

\[ d=\sqrt{4^2+3^2} \]

\[ d=\sqrt{16+9}=\sqrt{25} \]

Answer: \(d=5\) units

(ii) P(−1, 3), Q(3, −1)

Step 1: Identify the coordinates.

\[ x_1=-1,\quad y_1=3,\quad x_2=3,\quad y_2=-1 \]

Step 2: Apply the distance formula.

\[ d=\sqrt{(3-(-1))^2+(-1-3)^2} \]

\[ d=\sqrt{4^2+(-4)^2} \]

\[ d=\sqrt{16+16}=\sqrt{32} \]

Answer: \(d=4\sqrt{2}\) units

(iii) R(−3, −2), S(−7, −5)

Step 1: Identify the coordinates.

\[ x_1=-3,\quad y_1=-2,\quad x_2=-7,\quad y_2=-5 \]

Step 2: Apply the distance formula.

\[ d=\sqrt{(-7-(-3))^2+(-5-(-2))^2} \]

\[ d=\sqrt{(-4)^2+(-3)^2} \]

\[ d=\sqrt{16+9}=\sqrt{25} \]

Answer: \(d=5\) units

Q2. Find the midpoint of the line segment joining the following points.

Midpoint Formula

\[ M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right) \]

(i) A(1, 2), B(5, 8)

Step 1: Identify the coordinates.

\[ x_1=1,\quad y_1=2,\quad x_2=5,\quad y_2=8 \]

Step 2: Apply the midpoint formula.

\[ M=\left(\frac{1+5}{2},\frac{2+8}{2}\right) \]

\[ M=\left(\frac{6}{2},\frac{10}{2}\right) \]

Answer: \(M=(3,5)\)

(ii) P(−4, 6), Q(2, −2)

Step 1: Identify the coordinates.

\[ x_1=-4,\quad y_1=6,\quad x_2=2,\quad y_2=-2 \]

Step 2: Apply the midpoint formula.

\[ M=\left(\frac{-4+2}{2},\frac{6+(-2)}{2}\right) \]

\[ M=\left(\frac{-2}{2},\frac{4}{2}\right) \]

Answer: \(M=(-1,2)\)

(iii) R(−3, −5), S(−1, −1)

Step 1: Identify the coordinates.

\[ x_1=-3,\quad y_1=-5,\quad x_2=-1,\quad y_2=-1 \]

Step 2: Apply the midpoint formula.

\[ M=\left(\frac{-3+(-1)}{2},\frac{-5+(-1)}{2}\right) \]

\[ M=\left(\frac{-4}{2},\frac{-6}{2}\right) \]

Answer: \(M=(-2,-3)\)

Q3. Show that the points A(1, 1), B(4, 5), and C(7, 9) are collinear.

Step 1: Calculate the distance AB.

\[ AB=\sqrt{(4-1)^2+(5-1)^2} \]

\[ AB=\sqrt{3^2+4^2} \]

\[ AB=\sqrt{9+16}=\sqrt{25} \]

\[ AB=5 \]

Step 2: Calculate the distance BC.

\[ BC=\sqrt{(7-4)^2+(9-5)^2} \]

\[ BC=\sqrt{3^2+4^2} \]

\[ BC=\sqrt{9+16}=\sqrt{25} \]

\[ BC=5 \]

Step 3: Calculate the distance AC.

\[ AC=\sqrt{(7-1)^2+(9-1)^2} \]

\[ AC=\sqrt{6^2+8^2} \]

\[ AC=\sqrt{36+64}=\sqrt{100} \]

\[ AC=10 \]

Step 4: Verify collinearity.

\[ AB+BC=5+5=10=AC \]

Since \(AB+BC=AC\), the three points lie on the same straight line.

Hence proved: A, B, and C are collinear.

Q4. Show that the points A(0, 0), B(3, 0), and C(0, 4) form a right-angled triangle.

Step 1: Find the lengths of all three sides.

\[ AB=\sqrt{(3-0)^2+(0-0)^2} \]

\[ AB=\sqrt{9}=3 \]

\[ AC=\sqrt{(0-0)^2+(4-0)^2} \]

\[ AC=\sqrt{16}=4 \]

\[ BC=\sqrt{(0-3)^2+(4-0)^2} \]

\[ BC=\sqrt{(-3)^2+4^2} \]

\[ BC=\sqrt{9+16}=\sqrt{25}=5 \]

Step 2: Verify Pythagoras’ theorem.

\[ AB^2+AC^2=BC^2 \]

\[ 3^2+4^2=5^2 \]

\[ 9+16=25 \]

Since the side lengths satisfy Pythagoras’ theorem, triangle ABC is a right-angled triangle, with the right angle at A.

Hence proved.

Exercise 7.2

Q1. Find the slope (gradient) of the line passing through the following points.

Slope Formula

\[ m=\frac{y_2-y_1}{x_2-x_1} \]

(i) A(2, 3), B(5, 9)

\[ m=\frac{9-3}{5-2} \]

\[ m=\frac{6}{3} \]

Answer: \(m=2\)

(ii) P(−2, 4), Q(3, −1)

\[ m=\frac{-1-4}{3-(-2)} \]

\[ m=\frac{-5}{5} \]

Answer: \(m=-1\)

(iii) R(4, −2), S(4, 6)

\[ m=\frac{6-(-2)}{4-4} \]

\[ m=\frac{8}{0} \]

Division by zero is undefined.

Answer: The slope is undefined because the line is vertical.

Q2. Find the equation of the line passing through the point (2, 5) with slope \(m=3\).

Point-Slope Form

\[ y-y_1=m(x-x_1) \]

Step 1: Substitute the values.

\[ y-5=3(x-2) \]

Step 2: Simplify.

\[ y-5=3x-6 \]

\[ y=3x-1 \]

Answer: \(y=3x-1\)

The standard form is:

\[ 3x-y-1=0 \]

Q3. Find the equation of the line passing through the points A(1, 3) and B(4, 9).

Step 1: Find the slope.

\[ m=\frac{9-3}{4-1} \]

\[ m=\frac{6}{3}=2 \]

Step 2: Use the point-slope form with A(1, 3).

\[ y-3=2(x-1) \]

Step 3: Simplify.

\[ y-3=2x-2 \]

\[ y=2x+1 \]

Answer: \(y=2x+1\)

The standard form is:

\[ 2x-y+1=0 \]

Q4. Convert the linear equation \(3x+2y-6=0\) into slope-intercept form and state its slope and y-intercept.

Slope-Intercept Form

\[ y=mx+c \]

Step 1: Make \(y\) the subject.

\[ 3x+2y-6=0 \]

\[ 2y=-3x+6 \]

Divide both sides by 2:

\[ y=-\frac{3}{2}x+3 \]

Step 2: Compare with \(y=mx+c\).

Slope: \(m=-\frac{3}{2}\)
y-intercept: \(c=3\)
y-intercept point: \((0,3)\)

Answer:

\[ y=-\frac{3}{2}x+3 \]

Review Exercise 7

Q1. Multiple-Choice Questions (MCQs)

(i) The distance between the points (0, 0) and (3, 4) is:

Correct Answer: (a) 5

(ii) The midpoint of (2, 4) and (6, 8) is:

Correct Answer: (b) (4, 6)

(iii) The slope of a horizontal line is:

Correct Answer: (c) 0

(iv) The slope of a vertical line is:

Correct Answer: (b) Undefined

(v) If two lines are parallel, their slopes are:

Correct Answer: (a) Equal

(vi) The y-intercept of \(y=2x+5\) is:

Correct Answer: (c) 5

(vii) The distance formula between \((x_1,y_1)\) and \((x_2,y_2)\) is:

Correct Answer: (a)

\[ d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \]

Q2. Find the distance and midpoint between A(−3, 5) and B(3, −3).

Part (i): Distance

\[ d=\sqrt{(3-(-3))^2+(-3-5)^2} \]

\[ d=\sqrt{6^2+(-8)^2} \]

\[ d=\sqrt{36+64} \]

\[ d=\sqrt{100} \]

Answer: \(d=10\) units

Part (ii): Midpoint

\[ M=\left(\frac{-3+3}{2},\frac{5+(-3)}{2}\right) \]

\[ M=\left(\frac{0}{2},\frac{2}{2}\right) \]

Answer: \(M=(0,1)\)

Q3. Find the equation of the line passing through (3, −2) with slope \(m=-4\).

Step 1: Use the point-slope form.

\[ y-y_1=m(x-x_1) \]

Substitute \(x_1=3,\ y_1=-2,\ m=-4\):

\[ y-(-2)=-4(x-3) \]

\[ y+2=-4x+12 \]

Step 2: Simplify.

\[ y=-4x+10 \]

Answer: \(y=-4x+10\)

The standard form is:

\[ 4x+y-10=0 \]

Notes by Sir Dilawar

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