Rbse Solutions for Class 10 Maths Chapter 7 Exercise 7.2 Solutions | Section Formula

Rbse Solutions for Class 10 Maths Chapter 7 Exercise 7.2 Solutions | Section Formula

The problems in this exercise primarily use the Section Formula and the Midpoint Formula.

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

$$P(x, y) = \left(\frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}\right)$$

For the Midpoint, $m_1 = m_2 = 1$:

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

1. Find the coordinates of the point

Given points $A(-1, 7)$ and $B(4, -3)$, ratio $m_1: m_2 = 2: 3$.

$$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} = \frac{2(4) + 3(-1)}{2 + 3} = \frac{8 – 3}{5} = \frac{5}{5} = 1$$

$$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} = \frac{2(-3) + 3(7)}{2 + 3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3$$

The coordinates of the point are $\mathbf{(1, 3)}$.

2. Coordinates of the points of trisection

Points $A(4, -1)$ and $B(-2, -3)$. Trisection means dividing the segment into three equal parts by two points, $P$ and $Q$.

• Point $P$ divides $AB$ in the ratio $1: 2$ ($m_1=1, m_2=2$).$$x_P = \frac{1(-2) + 2(4)}{1 + 2} = \frac{-2 + 8}{3} = \frac{6}{3} = 2$$$$y_P = \frac{1(-3) + 2(-1)}{1 + 2} = \frac{-3 – 2}{3} = -\frac{5}{3}$$$P$ is $\mathbf{(2, -5/3)}$.
• Point $Q$ divides $AB$ in the ratio $2: 1$ ($m_1=2, m_2=1$).$$x_Q = \frac{2(-2) + 1(4)}{2 + 1} = \frac{-4 + 4}{3} = 0$$$$y_Q = \frac{2(-3) + 1(-1)}{2 + 1} = \frac{-6 – 1}{3} = -\frac{7}{3}$$$Q$ is $\mathbf{(0, -7/3)}$.

The coordinates of the points of trisection are $\mathbf{(2, -5/3)}$ and $\mathbf{(0, -7/3)}$.

3. Sports Day Activities

Total distance along AD is 100 flower pots $\times$ 1m/pot = 100m.

1. Niharika’s Flag (Green):
   • Line number (x-coordinate): $x_N = 2$.
   • Distance along AD (y-coordinate): $y_N = \frac{1}{4} \times 100 \text{m} = 25 \text{m}$.
   • Niharika’s flag is at $N(2, 25)$.
2. Preet’s Flag (Red):
   • Line number (x-coordinate): $x_P = 8$.
   • Distance along AD (y-coordinate): $y_P = \frac{1}{5} \times 100 \text{m} = 20 \text{m}$.
   • Preet’s flag is at $P(8, 20)$.
3. Distance between the two flags ($NP$):$$NP = \sqrt{(8 – 2)^2 + (20 – 25)^2} = \sqrt{6^2 + (-5)^2} = \sqrt{36 + 25} = \sqrt{61}$$The distance between both flags is $\mathbf{\sqrt{61} \text{ meters}}$.
4. Rashmi’s Flag (Blue):Rashmi posts her flag exactly halfway, which is the midpoint of $NP$.$$R(x_R, y_R) = \left(\frac{2 + 8}{2}, \frac{25 + 20}{2}\right) = \left(\frac{10}{2}, \frac{45}{2}\right) = (5, 22.5)$$Rashmi should post her flag on the $5^{\text{th}}$ line at a distance of $22.5 \text{ meters}$ from the starting point along AD.

4. Find the ratio of division

Let the point $P(-1, 6)$ divide the line segment joining $A(-3, 10)$ and $B(6, -8)$ in the ratio $k: 1$.

Using the x-coordinate:

$$x = \frac{k x_2 + 1 x_1}{k + 1}$$

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

$$-1(k + 1) = 6k – 3$$

$$-k – 1 = 6k – 3$$

$$3 – 1 = 6k + k$$

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

The ratio is $k: 1 = \mathbf{2: 7}$. (You can verify with the y-coordinate: $6 = \frac{2(-8) + 7(10)}{2+7} = \frac{-16 + 70}{9} = \frac{54}{9} = 6$).

5. Division by the x-axis

The x-axis is the line $y=0$. A point on the x-axis has coordinates $P(x, 0)$.

Let the x-axis divide $A(1, -5)$ and $B(-4, 5)$ in the ratio $k: 1$.

1. Find the ratio ($k$): Use the y-coordinate, which is known (0).$$y = \frac{k y_2 + 1 y_1}{k + 1}$$$$0 = \frac{k(5) + 1(-5)}{k + 1}$$$$0 = 5k – 5 \implies 5k = 5 \implies k = 1$$The ratio is $k: 1 = \mathbf{1: 1}$. (This means the x-axis is the perpendicular bisector and the point of division is the midpoint.)
2. Find the coordinates of the point of division ($P$):$$x = \frac{1(-4) + 1(1)}{1 + 1} = \frac{-3}{2}$$$$y = 0$$The coordinates of the point of division are $\mathbf{(-3/2, 0)}$.

6. Vertices of a Parallelogram

The vertices are $A(1, 2), B(4, y), C(x, 6), D(3, 5)$ in order.

In a parallelogram, the diagonals bisect each other. Therefore, the midpoint of diagonal $AC$ must be the same as the midpoint of diagonal $BD$.

1. Midpoint of $AC$:$$M_{AC} = \left(\frac{1 + x}{2}, \frac{2 + 6}{2}\right) = \left(\frac{1 + x}{2}, 4\right)$$
2. Midpoint of $BD$:$$M_{BD} = \left(\frac{4 + 3}{2}, \frac{y + 5}{2}\right) = \left(\frac{7}{2}, \frac{y + 5}{2}\right)$$
3. Equate the coordinates:
   • x-coordinates:$$\frac{1 + x}{2} = \frac{7}{2} \implies 1 + x = 7 \implies \mathbf{x = 6}$$
   • y-coordinates:$$4 = \frac{y + 5}{2} \implies 8 = y + 5 \implies \mathbf{y = 3}$$

7. Coordinates of point A (Diameter)

$AB$ is the diameter, and $C(2, -3)$ is the center (midpoint of $AB$). $B$ is $(1, 4)$. Let $A$ be $(x, y)$.

Using the Midpoint Formula for $C$:

$$C_x = \frac{x_A + x_B}{2} \quad \implies \quad 2 = \frac{x + 1}{2}$$

$$C_y = \frac{y_A + y_B}{2} \quad \implies \quad -3 = \frac{y + 4}{2}$$

Solve for $x$:

$$4 = x + 1 \implies x = 3$$

Solve for $y$:

$$-6 = y + 4 \implies y = -10$$

The coordinates of point A are $\mathbf{(3, -10)}$.

8. Find coordinates of P

Given $A(-2, -2)$ and $B(2, -4)$, and $AP = \frac{3}{7} AB$.

If $AP = \frac{3}{7} AB$, then $PB = AB – AP = AB – \frac{3}{7} AB = \frac{4}{7} AB$.

The point $P$ divides $AB$ in the ratio $AP: PB = \frac{3}{7} AB : \frac{4}{7} AB = 3: 4$.

So, $m_1=3, m_2=4$.

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

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

The coordinates of P are $\mathbf{(-2/7, -20/7)}$.

9. Divide the segment into four equal parts

Points $A(-2, 2)$ and $B(2, 8)$. Let the points be $P_1, P_2, P_3$.

1. Point $P_2$ is the midpoint of $AB$ (ratio $1: 1$).$$P_2 = \left(\frac{-2 + 2}{2}, \frac{2 + 8}{2}\right) = (0, 5)$$
2. Point $P_1$ is the midpoint of $AP_2$ (ratio $1: 1$).$$P_1 = \left(\frac{-2 + 0}{2}, \frac{2 + 5}{2}\right) = \left(-1, \frac{7}{2}\right)$$
3. Point $P_3$ is the midpoint of $P_2 B$ (ratio $1: 1$).$$P_3 = \left(\frac{0 + 2}{2}, \frac{5 + 8}{2}\right) = \left(1, \frac{13}{2}\right)$$

The coordinates of the points dividing $AB$ into four equal parts are $\mathbf{(-1, 7/2), (0, 5), \text{ and } (1, 13/2)}$.

10. Area of a Rhombus

Vertices $A(3, 0), B(4, 5), C(-1, 4), D(-2, -1)$.

Area $= \frac{1}{2} \times (\text{product of diagonals}) = \frac{1}{2} \times (AC \times BD)$.

1. Length of Diagonal $AC$:$$AC = \sqrt{(-1 – 3)^2 + (4 – 0)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$$
2. Length of Diagonal $BD$:$$BD = \sqrt{(-2 – 4)^2 + (-1 – 5)^2} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$$
3. Area of Rhombus:$$\text{Area} = \frac{1}{2} \times AC \times BD = \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2}$$$$\text{Area} = \frac{1}{2} \times (4 \times 6) \times (\sqrt{2} \times \sqrt{2})$$$$\text{Area} = \frac{1}{2} \times 24 \times 2 = \mathbf{24 \text{ square units}}$$

Last Updated on November 28, 2025 by Aman Singh