Class 10 Maths Chapter 7 Exercise 7.2 NCERT Solutions

Question 1 NCERT Ex 7.2

Find the coordinates of the point which divides the join of $(-2, 5)$ and $(3, -5)$ in the ratio $2 : 3$.

Step 1: Identify Given Values

Points: $A(x_1, y_1) = (-2, 5)$ and $B(x_2, y_2) = (3, -5)$.
Ratio: $m_1 : m_2 = 2 : 3$.

Step 2: Apply Section Formula

$$x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} = \frac{2(3) + 3(-2)}{2 + 3} = \frac{6 - 6}{5} = \frac{0}{5} = 0$$ $$y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} = \frac{2(-5) + 3(5)}{2 + 3} = \frac{-10 + 15}{5} = \frac{5}{5} = 1$$

Required coordinates of the point are $(0, 1)$.

Question 2 NCERT Ex 7.2

Find the coordinates of the points of trisection of the line segment joining $(6, -2)$ and $(-3, 4)$.

Step 1: Understanding Points of Trisection

Points of trisection divide segment $AB$ into 3 equal parts: $AP = PQ = QB$.
Therefore, point $P$ divides $AB$ in the ratio $1 : 2$, and point $Q$ divides $AB$ in the ratio $2 : 1$ (or is the midpoint of $PB$).
Given: $A(6, -2)$ and $B(-3, 4)$.

Step 2: Find Coordinates of Point $P$ (Ratio $1 : 2$)

$$x_P = \frac{1(-3) + 2(6)}{1 + 2} = \frac{-3 + 12}{3} = \frac{9}{3} = 3$$ $$y_P = \frac{1(4) + 2(-2)}{1 + 2} = \frac{4 - 4}{3} = \frac{0}{3} = 0$$ Thus, $P(3, 0)$.

Step 3: Find Coordinates of Point $Q$ (Ratio $2 : 1$)

$$x_Q = \frac{2(-3) + 1(6)}{2 + 1} = \frac{-6 + 6}{3} = \frac{0}{3} = 0$$ $$y_Q = \frac{2(4) + 1(-2)}{2 + 1} = \frac{8 - 2}{3} = \frac{6}{3} = 2$$ Thus, $Q(0, 2)$.

The coordinates of the points of trisection are $(3, 0)$ and $(0, 2)$.

Question 3 NCERT Ex 7.2

To conduct Sports Day activities in a rectangular shaped school ground $ABCD$, lines have been drawn with chalk powder at a distance of $1\text{ m}$ each. $80$ flower pots have been placed at a distance of $1\text{ m}$ from each other along $AD$. Student A runs $\frac{1}{4}\text{th}$ the distance $AD$ on the $2\text{nd}$ line and posts a green flag. Student B runs $\frac{1}{5}\text{th}$ the distance $AD$ on the $8\text{th}$ line and posts a red flag. What is the distance between both flags? If Student C has to post a blue flag exactly halfway between the line segment joining the two flags, where should it be posted?

Step 1: Determine Coordinates of the Flags

Total distance along $AD = 80\text{ m}$.
Green Flag ($G$): Line $2$, distance $= \frac{1}{4} \times 80 = 20\text{ m} \implies G(2, 20)$.
Red Flag ($R$): Line $8$, distance $= \frac{1}{5} \times 80 = 16\text{ m} \implies R(8, 16)$.

Step 2: Distance Between the Two Flags

Using distance formula: $$GR = \sqrt{(8 - 2)^2 + (16 - 20)^2} = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}\text{ m} \approx 7.21\text{ m}$$

Step 3: Midpoint Position for the Blue Flag

Coordinates of midpoint $M$: $$x = \frac{2 + 8}{2} = \frac{10}{2} = 5$$ $$y = \frac{20 + 16}{2} = \frac{36}{2} = 18$$ The blue flag should be placed at coordinates $(5, 18)$.

Distance between flags = $\sqrt{52} = 2\sqrt{13}\text{ m}$ (approx $7.21\text{ m}$); Blue flag is on the $5\text{th}$ line at a distance of $18\text{ m}$.

Question 4 NCERT Ex 7.2

Find the ratio in which the line segment joining the points $(-2, 5)$ and $(4, -4)$ is divided by the point $(-1, 3.5)$.

Step 1: Assume the Ratio as $k : 1$

Let the required ratio be $k : 1$.
Given endpoints $A(-2, 5)$ and $B(4, -4)$, and division point $P(-1, 3.5)$.
Using section formula for the $x$-coordinate: $$x = \frac{k(x_2) + 1(x_1)}{k + 1}$$ $$-1 = \frac{k(4) + 1(-2)}{k + 1}$$

Step 2: Solve for $k$

$$-1(k + 1) = 4k - 2$$ $$-k - 1 = 4k - 2$$ $$-k - 4k = -2 + 1$$ $$-5k = -1 \implies k = \frac{1}{5}$$ Therefore, the ratio is $\frac{1}{5} : 1$, which is $1 : 5$.

Step 3: Verification with $y$-coordinate

$$y = \frac{1(-4) + 5(5)}{1 + 5} = \frac{-4 + 25}{6} = \frac{21}{6} = 3.5$$ This matches the given $y$-coordinate ($3.5$).

The required ratio is $1 : 5$.

Question 5 NCERT Ex 7.2

Find the ratio in which the line segment joining $A(2, -4)$ and $B(-3, 6)$ is divided by the $x$-axis. Also find the coordinates of the point of division.

Step 1: Property of Point on $x$-axis

Any point lying on the $x$-axis has its $y$-coordinate equal to $0$.
Let the point of division be $P(x, 0)$ and the ratio be $k : 1$.

Step 2: Use $y$-coordinate to Find $k$

$$y = \frac{k(y_2) + 1(y_1)}{k + 1}$$ $$0 = \frac{k(6) + 1(-4)}{k + 1}$$ $$6k - 4 = 0 \implies 6k = 4 \implies k = \frac{4}{6} = \frac{2}{3}$$ Thus, the ratio is $2 : 3$.

Step 3: Find the $x$-coordinate

Substitute $m_1 = 2$ and $m_2 = 3$: $$x = \frac{2(-3) + 3(2)}{2 + 3} = \frac{-6 + 6}{5} = \frac{0}{5} = 0$$ Therefore, the point of division is $(0, 0)$.

Ratio is $2 : 3$ and the point of division is $(0, 0)$.

Question 6 NCERT Ex 7.2

If $(2, 3)$, $(5, y)$, $(x, 7)$ and $(4, 6)$ are the vertices of a parallelogram taken in order, find $x$ and $y$.

Step 1: Geometric Property of Parallelograms

Let $A(2, 3), B(5, y), C(x, 7)$, and $D(4, 6)$ be the vertices taken in order.
A fundamental property of a parallelogram is that its diagonals bisect each other.
Therefore, Midpoint of diagonal $AC$ = Midpoint of diagonal $BD$.

Step 2: Compute and Equate Coordinates

Midpoint of $AC$: $$\left(\frac{2 + x}{2}, \; \frac{3 + 7}{2}\right) = \left(\frac{2 + x}{2}, \; 5\right)$$ Midpoint of $BD$: $$\left(\frac{5 + 4}{2}, \; \frac{y + 6}{2}\right) = \left(\frac{9}{2}, \; \frac{y + 6}{2}\right)$$ Equating the $x$-coordinates: $$\frac{2 + x}{2} = \frac{9}{2} \implies 2 + x = 9 \implies x = 7$$ Equating the $y$-coordinates: $$\frac{y + 6}{2} = 5 \implies y + 6 = 10 \implies y = 4$$

$x = 7$ and $y = 4$.

Question 7 NCERT Ex 7.2

Find the coordinates of point $A$, where $AB$ is the diameter of a circle whose centre is $(3, -2)$ and $B$ is $(2, 5)$.

Step 1: Use the Midpoint Formula

The centre of a circle is always the midpoint of its diameter.
Let $A$ have coordinates $(x, y)$.
Given centre $C(3, -2)$ and endpoint $B(2, 5)$.
Using the midpoint formula: $$x_{\text{mid}} = \frac{x_A + x_B}{2} \implies 3 = \frac{x + 2}{2}$$ $$y_{\text{mid}} = \frac{y_A + y_B}{2} \implies -2 = \frac{y + 5}{2}$$

Step 2: Solve for $(x, y)$

$$3 = \frac{x + 2}{2} \implies x + 2 = 6 \implies x = 6 - 2 = 4$$ $$-2 = \frac{y + 5}{2} \implies y + 5 = -4 \implies y = -4 - 5 = -9$$

The coordinates of point $A$ are $(4, -9)$.

Question 8 NCERT Ex 7.2

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

Step 1: Determine the Ratio $AP : PB$

Given $AP = \frac{4}{7}AB$.
Since $P$ lies on segment $AB$: $$PB = AB - AP = AB - \frac{4}{7}AB = \frac{3}{7}AB$$ Therefore: $$\frac{AP}{PB} = \frac{\frac{4}{7}AB}{\frac{3}{7}AB} = \frac{4}{3} \implies AP : PB = 4 : 3$$

Step 2: Apply Section Formula ($4 : 3$)

Here $(x_1, y_1) = (-2, -2)$, $(x_2, y_2) = (2, -4)$, $m_1 = 4$, $m_2 = 3$.
$$x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} = \frac{4(2) + 3(-2)}{4 + 3} = \frac{8 - 6}{7} = \frac{2}{7}$$ $$y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} = \frac{4(-4) + 3(-2)}{4 + 3} = \frac{-16 - 6}{7} = -\frac{22}{7}$$

The coordinates of $P$ are $\left(\frac{2}{7}, -\frac{22}{7}\right)$.

Question 9 NCERT Ex 7.2

Find the coordinates of the points which divide the line segment joining $A(-4, 0)$ and $B(4, 8)$ into four equal parts.

Step 1: Geometric Strategy Using Midpoints

Let the points dividing $AB$ into 4 equal segments be $P_1, P_2$, and $P_3$.
- $P_2$ is the midpoint of $AB$.
- $P_1$ is the midpoint of $AP_2$.
- $P_3$ is the midpoint of $P_2B$.

Step 2: Compute Midpoint $P_2$

$$P_2 = \left(\frac{-4 + 4}{2}, \; \frac{0 + 8}{2}\right) = (0, 4)$$

Step 3: Compute Midpoint $P_1$ (Between $A(-4, 0)$ and $P_2(0, 4)$)

$$P_1 = \left(\frac{-4 + 0}{2}, \; \frac{0 + 4}{2}\right) = (-2, 2)$$

Step 4: Compute Midpoint $P_3$ (Between $P_2(0, 4)$ and $B(4, 8)$)

$$P_3 = \left(\frac{0 + 4}{2}, \; \frac{4 + 8}{2}\right) = (2, 6)$$

The coordinates of the three division points are $(-2, 2)$, $(0, 4)$, and $(2, 6)$.

Question 10 NCERT Ex 7.2

Find the area of a rhombus if its vertices are $(4, 0)$, $(5, 6)$, $(-2, 5)$ and $(-3, -1)$ taken in order.

Step 1: Area Formula of a Rhombus

$$\text{Area of Rhombus} = \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times (AC) \times (BD)$$ Let $A(4, 0), B(5, 6), C(-2, 5)$, and $D(-3, -1)$.

Step 2: Compute Lengths of Diagonals $AC$ and $BD$

Diagonal $AC$: $$AC = \sqrt{(-2 - 4)^2 + (5 - 0)^2} = \sqrt{(-6)^2 + 5^2} = \sqrt{36 + 25} = \sqrt{61}$$ Diagonal $BD$: $$BD = \sqrt{(-3 - 5)^2 + (-1 - 6)^2} = \sqrt{(-8)^2 + (-7)^2} = \sqrt{64 + 49} = \sqrt{113}$$

Step 3: Calculate the Area

$$\text{Area} = \frac{1}{2} \times \sqrt{61} \times \sqrt{113} = \frac{1}{2}\sqrt{6893} \approx \frac{1}{2} \times 83.024 = 41.51\text{ sq units}$$

Area of rhombus = $\frac{1}{2}\sqrt{6893}\text{ sq units}$ (approx $41.51\text{ sq units}$).

Coordinate Geometry - Exercise 7.2

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