NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.2#
Exercise 7.2 covers the Section Formula — finding the coordinates of a point that divides a line segment in a given ratio, and its special case (midpoint formula).
Section Formula: The coordinates of point $P$ which divides the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ in ratio $m:n$ internally are:
$$P = \left(\frac{mx_2 + nx_1}{m+n},\ \frac{my_2 + ny_1}{m+n}\right)$$Midpoint Formula: Midpoint of AB $= \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$
1. Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3. #
Answer
Let $A(-1, 7)$ and $B(4, -3)$, with $m : n = 2 : 3$.
Using the section formula, we find the $x$-coordinate:
$$= \frac{2 \times 4 + 3 \times (-1)}{2 + 3}$$$$= \frac{8 - 3}{5}$$
$$= 1$$
Using the section formula, we find the $y$-coordinate:
$$= \frac{2 \times (-3) + 3 \times 7}{2 + 3}$$$$= \frac{-6 + 21}{5}$$
$$= 3$$
∴ The required point is $\mathbf{(1, 3)}$.
2. Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3). #
Answer
Let $A(4, -1)$ and $B(-2, -3)$. The points of trisection P and Q divide AB in the ratios 1 : 2 and 2 : 1.
Point P (ratio 1 : 2):
$$x = \frac{1 \times (-2) + 2 \times 4}{3} = \frac{6}{3} = 2$$$$y = \frac{1 \times (-3) + 2 \times (-1)}{3} = -\frac{5}{3}$$
Point Q (ratio 2 : 1):
$$x = \frac{2 \times (-2) + 1 \times 4}{3} = \frac{0}{3} = 0$$$$y = \frac{2 \times (-3) + 1 \times (-1)}{3} = -\frac{7}{3}$$
∴ The points of trisection are $\mathbf{P\left(2, -\dfrac{5}{3}\right)}$ and $\mathbf{Q\left(0, -\dfrac{7}{3}\right)}$.
3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD as shown in the figure. Niharika runs $\frac{1}{4}$th the distance AD on the 2nd line and posts a green flag. Preet runs $\frac{1}{5}$th the distance AD on the eighth 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? #
Answer
Take A as the origin, AB along the x-axis and AD along the y-axis.
There are 100 flower pots 1 m apart along AD, so AD = 100 m.
Niharika runs $\dfrac{1}{4}$ of AD on the 2nd line:
Distance run = $\dfrac{1}{4} \times 100 = 25$ m
Green flag: $G(2, 25)$
Preet runs $\dfrac{1}{5}$ of AD on the 8th line:
Distance run = $\dfrac{1}{5} \times 100 = 20$ m
Red flag: $R(8, 20)$
Distance between the flags: Using the distance formula, we find GR:
$$= \sqrt{(8-2)^2 + (20-25)^2}$$$$= \sqrt{36 + 25}$$
$$= \sqrt{61}$$
Blue flag: It is at the midpoint of GR.
$$x = \frac{2 + 8}{2} = 5$$$$y = \frac{25 + 20}{2} = 22.5$$
∴ The distance between the flags is $\mathbf{\sqrt{61}}$ m, and Rashmi should post the blue flag on the 5th line, 22.5 m from AB, at (5, 22.5).
4. Find the ratio in which the line segment joining the points (–3, 10) and (6, –8) is divided by (–1, 6). #
Answer
Let the point $(-1, 6)$ divide the segment joining $A(-3, 10)$ and $B(6, -8)$ in the ratio $k : 1$.
Using the section formula, the x-coordinate is −1, so:
$$\frac{6k + (-3)}{k + 1} = -1$$$$\Rightarrow 6k - 3 = -k - 1$$
$$\Rightarrow 7k = 2$$
$$\Rightarrow k = \frac{2}{7}$$
Check with the y-coordinate. For $k = \dfrac{2}{7}$, we find $\dfrac{-8k + 10}{k + 1}$:
$$= \frac{-\frac{16}{7} + 10}{\frac{2}{7} + 1}$$$$= \frac{\frac{54}{7}}{\frac{9}{7}}$$
$$= 6$$
This matches the y-coordinate of $(-1, 6)$.
∴ The ratio is $\dfrac{2}{7} : 1$, that is 2 : 7.
5. 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. #
Answer
Let the x-axis divide AB in the ratio $k : 1$ at the point $P(x, 0)$.
P lies on the x-axis, so its y-coordinate is 0:
$$\frac{5k + (-5)}{k + 1} = 0$$$$\Rightarrow 5k - 5 = 0$$
$$\Rightarrow k = 1$$
So the ratio is 1 : 1, and P is the midpoint of AB:
$$x = \frac{1 \times (-4) + 1 \times 1}{1 + 1} = -\frac{3}{2}$$∴ The x-axis divides AB in the ratio 1 : 1 at the point $\mathbf{\left(-\dfrac{3}{2}, 0\right)}$.
6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y. #
Answer
Let $A(1, 2)$, $B(4, y)$, $C(x, 6)$ and $D(3, 5)$ be the vertices of parallelogram ABCD.
The diagonals of a parallelogram bisect each other, so the midpoint of AC is the same as the midpoint of BD.
Midpoint of AC:
$$\left(\frac{1 + x}{2}, \frac{2 + 6}{2}\right) = \left(\frac{1 + x}{2}, 4\right)$$Midpoint of BD:
$$\left(\frac{4 + 3}{2}, \frac{y + 5}{2}\right) = \left(\frac{7}{2}, \frac{y + 5}{2}\right)$$The x-coordinates are equal, so:
$$\frac{1 + x}{2} = \frac{7}{2}$$$$\Rightarrow 1 + x = 7$$
$$\Rightarrow x = 6$$
The y-coordinates are equal, so:
$$\frac{y + 5}{2} = 4$$$$\Rightarrow y + 5 = 8$$
$$\Rightarrow y = 3$$
∴ $x = \mathbf{6}$ and $y = \mathbf{3}$
7. Find the coordinates of point A, where AB is the diameter of a circle whose centre is (2, –3) and B is (1, 4). #
Answer
Let the point be $A(x, y)$. The centre $(2, -3)$ is the midpoint of the diameter AB.
The x-coordinate of the midpoint is 2, so:
$$\frac{x + 1}{2} = 2$$$$\Rightarrow x + 1 = 4$$
$$\Rightarrow x = 3$$
The y-coordinate of the midpoint is −3, so:
$$\frac{y + 4}{2} = -3$$$$\Rightarrow y + 4 = -6$$
$$\Rightarrow y = -10$$
∴ $A = \mathbf{(3, -10)}$
8. If A and B are (–2, –2) and (2, –4), respectively, find the coordinates of P such that AP = $\dfrac{3}{7}$AB and P lies on the line segment AB. #
Answer
P lies on AB and $AP = \dfrac{3}{7}AB$, so:
$PB = AB - \dfrac{3}{7}AB = \dfrac{4}{7}AB$
$AP : PB = 3 : 4$
Using the section formula with $A(-2, -2)$, $B(2, -4)$ and $m : n = 3 : 4$:
$$x = \frac{3 \times 2 + 4 \times (-2)}{3 + 4} = -\frac{2}{7}$$$$y = \frac{3 \times (-4) + 4 \times (-2)}{3 + 4} = -\frac{20}{7}$$
∴ $P = \mathbf{\left(-\dfrac{2}{7}, -\dfrac{20}{7}\right)}$
9. Find the coordinates of the points which divide the line segment joining A(–2, 2) and B(2, 8) into four equal parts. #
Answer
Let $P_1$, $P_2$ and $P_3$ divide AB into four equal parts. They divide AB in the ratios 1 : 3, 1 : 1 and 3 : 1.
$P_1$ (ratio 1 : 3):
$$x = \frac{1 \times 2 + 3 \times (-2)}{4} = -1$$$$y = \frac{1 \times 8 + 3 \times 2}{4} = \frac{7}{2}$$
$P_2$ (midpoint):
$$x = \frac{-2 + 2}{2} = 0$$$$y = \frac{2 + 8}{2} = 5$$
$P_3$ (ratio 3 : 1):
$$x = \frac{3 \times 2 + 1 \times (-2)}{4} = 1$$$$y = \frac{3 \times 8 + 1 \times 2}{4} = \frac{13}{2}$$
∴ The points are $\mathbf{\left(-1, \dfrac{7}{2}\right)}$, $\mathbf{(0, 5)}$ and $\mathbf{\left(1, \dfrac{13}{2}\right)}$.
10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (–1, 4) and (–2, –1) taken in order. (Hint: Area of a rhombus = $\dfrac{1}{2}$ × d₁ × d₂) #
Answer
Let the vertices be $A(3, 0)$, $B(4, 5)$, $C(-1, 4)$ and $D(-2, -1)$.
Diagonals:
$$d_1 = AC = \sqrt{(-1-3)^2 + (4-0)^2} = \sqrt{32} = 4\sqrt{2}$$$$d_2 = BD = \sqrt{(-2-4)^2 + (-1-5)^2} = \sqrt{72} = 6\sqrt{2}$$
Using Area $= \dfrac{1}{2} \times d_1 \times d_2$, we find the area of the rhombus:
$$= \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2}$$$$= \frac{1}{2} \times 48$$
$$= 24$$
∴ The area of the rhombus is 24 square units.