NCERT Class 10 Maths – Exercise 7.2 Solutions

NCERT Class 10 Maths

Chapter 7 – Coordinate Geometry | Exercise 7.2

(Rationalized Syllabus 2025-26)

💡 Key Formulas

Section Formula: The coordinates of point $P(x, y)$ dividing the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ in the ratio $m_1 : m_2$ are:

$$P(x, y) = \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right)$$

Midpoint Formula: (Ratio 1:1)

$$M(x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$$

Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.

Let $A(-1, 7)$ and $B(4, -3)$. Ratio $m_1:m_2 = 2:3$.

Using Section Formula:

$$x = \frac{2(4) + 3(-1)}{2+3} = \frac{8 – 3}{5} = \frac{5}{5} = 1$$

$$y = \frac{2(-3) + 3(7)}{2+3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3$$

✔️ Coordinates: (1, 3)

Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).

Let $A(4, -1)$ and $B(-2, -3)$.

Points of trisection $P$ and $Q$ divide AB in ratios 1:2 and 2:1 respectively.

For point P (1:2):

$$x_p = \frac{1(-2) + 2(4)}{1+2} = \frac{-2 + 8}{3} = 2$$

$$y_p = \frac{1(-3) + 2(-1)}{1+2} = \frac{-3 – 2}{3} = -\frac{5}{3}$$

For point Q (2:1):

$$x_q = \frac{2(-2) + 1(4)}{2+1} = \frac{-4 + 4}{3} = 0$$

$$y_q = \frac{2(-3) + 1(-1)}{2+1} = \frac{-6 – 1}{3} = -\frac{7}{3}$$

✔️ P(2, -5/3), Q(0, -7/3)

To conduct Sports Day activities, lines have been drawn with chalk powder at a distance of 1m each. Niharika runs 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th the distance AD on the 8th line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

Total distance AD = 100m.

Niharika’s position (Green Flag): $x_1 = 2$, $y_1 = \frac{1}{4} \times 100 = 25$. So, $G(2, 25)$.

Preet’s position (Red Flag): $x_2 = 8$, $y_2 = \frac{1}{5} \times 100 = 20$. So, $R(8, 20)$.

Distance between flags:

$$GR = \sqrt{(8-2)^2 + (20-25)^2} = \sqrt{36 + 25} = \sqrt{61} \text{ m}$$

Rashmi’s position (Midpoint):

$$x = \frac{2+8}{2} = 5, \quad y = \frac{25+20}{2} = 22.5$$

✔️ Distance = $\sqrt{61}$ m, Blue Flag at (5, 22.5)

Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).

Let the ratio be $k:1$. Using the x-coordinate of the dividing point $P(-1, 6)$:

$$-1 = \frac{k(6) + 1(-3)}{k+1}$$

$$-k – 1 = 6k – 3$$

$$2 = 7k \Rightarrow k = \frac{2}{7}$$

The ratio is $2/7 : 1$ or $2:7$.

✔️ Ratio is 2:7

Find the ratio in which the line segment joining A(1, -5) and B(-4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Let the ratio be $k:1$. A point on the x-axis has coordinates $(x, 0)$.

Using the y-coordinate formula (since y=0):

$$0 = \frac{k(5) + 1(-5)}{k+1}$$

$$5k – 5 = 0 \Rightarrow 5k = 5 \Rightarrow k = 1$$

Ratio is 1:1 (Midpoint).

Now find x-coordinate:

$$x = \frac{1(-4) + 1(1)}{2} = \frac{-3}{2}$$

✔️ Ratio 1:1, Point (-3/2, 0)

If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

The diagonals of a parallelogram bisect each other. So, midpoint of AC = midpoint of BD.

Midpoint of AC:

$$O = \left( \frac{1+x}{2}, \frac{2+6}{2} \right) = \left( \frac{1+x}{2}, 4 \right)$$

Midpoint of BD:

$$O = \left( \frac{4+3}{2}, \frac{y+5}{2} \right) = \left( \frac{7}{2}, \frac{y+5}{2} \right)$$

Equating coordinates:

$$\frac{1+x}{2} = \frac{7}{2} \Rightarrow 1+x = 7 \Rightarrow x = 6$$

$$4 = \frac{y+5}{2} \Rightarrow 8 = y+5 \Rightarrow y = 3$$

✔️ x = 6, y = 3

Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).

Let $A(x, y)$. Center $C(2, -3)$ is the midpoint of AB.

$$\frac{x + 1}{2} = 2 \Rightarrow x + 1 = 4 \Rightarrow x = 3$$

$$\frac{y + 4}{2} = -3 \Rightarrow y + 4 = -6 \Rightarrow y = -10$$

✔️ A(3, -10)

If A and B are (-2, -2) and (2, -4), find the coordinates of P such that $AP = \frac{3}{7}AB$ and P lies on the line segment AB.

Given $AP = \frac{3}{7}AB \Rightarrow \frac{AP}{AB} = \frac{3}{7}$.

Since $AB = AP + PB$, ratio $AP:PB = 3:(7-3) = 3:4$.

Using Section Formula with $m_1=3, m_2=4$:

$$x = \frac{3(2) + 4(-2)}{3+4} = \frac{6 – 8}{7} = -\frac{2}{7}$$

$$y = \frac{3(-4) + 4(-2)}{3+4} = \frac{-12 – 8}{7} = -\frac{20}{7}$$

✔️ P(-2/7, -20/7)

Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.

Let points be $P_1, P_2, P_3$. $P_2$ is the midpoint of AB.

$$P_2 = \left( \frac{-2+2}{2}, \frac{2+8}{2} \right) = (0, 5)$$

$P_1$ is the midpoint of $A P_2$:

$$P_1 = \left( \frac{-2+0}{2}, \frac{2+5}{2} \right) = (-1, 3.5)$$

$P_3$ is the midpoint of $P_2 B$:

$$P_3 = \left( \frac{0+2}{2}, \frac{5+8}{2} \right) = (1, 6.5)$$

✔️ (-1, 7/2), (0, 5), (1, 13/2)

Q10

Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.

Area of Rhombus = $\frac{1}{2} \times d_1 \times d_2$.

Diagonal AC: $A(3, 0), C(-1, 4)$.

$$d_1 = \sqrt{(-1-3)^2 + (4-0)^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}$$

Diagonal BD: $B(4, 5), D(-2, -1)$.

$$d_2 = \sqrt{(-2-4)^2 + (-1-5)^2} = \sqrt{36+36} = \sqrt{72} = 6\sqrt{2}$$

Area:

$$\text{Area} = \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = \frac{1}{2} \times 24 \times 2 = 24$$

✔️ 24 square units

🎉 Exercise 7.2 Completed