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 $A(-1, 7)$, $B(4, -3)$, ratio $m:n = 2:3$.
$$x = \frac{2 \times 4 + 3 \times (-1)}{2+3} = \frac{8-3}{5} = \frac{5}{5} = 1$$$$y = \frac{2 \times (-3) + 3 \times 7}{2+3} = \frac{-6+21}{5} = \frac{15}{5} = 3$$∴ 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 trisection points $P$ and $Q$ divide $AB$ in ratios $1:2$ and $2:1$.
Point P (ratio 1:2):
$$P_x = \frac{1 \times (-2) + 2 \times 4}{3} = \frac{-2+8}{3} = 2$$$$P_y = \frac{1 \times (-3) + 2 \times (-1)}{3} = \frac{-3-2}{3} = \frac{-5}{3}$$Point Q (ratio 2:1):
$$Q_x = \frac{2 \times (-2) + 1 \times 4}{3} = \frac{-4+4}{3} = 0$$$$Q_y = \frac{2 \times (-3) + 1 \times (-1)}{3} = \frac{-6-1}{3} = \frac{-7}{3}$$∴ The trisection points 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
Let the school ground be set up with x-axis along AB.
AD = 100 m (100 flower pots at 1 m each).
Niharika: On 2nd line (x = 2), runs $\dfrac{1}{4}$ of AD = 25 m. Green flag position: $G(2, 25)$
Preet: On 8th line (x = 8), runs $\dfrac{1}{5}$ of AD = 20 m. Red flag position: $R(8, 20)$
Distance between flags:
$$GR = \sqrt{(8-2)^2 + (20-25)^2} = \sqrt{36+25} = \sqrt{61} \approx 7.81 \text{ m}$$Rashmi’s blue flag (midpoint):
$$x = \frac{2+8}{2} = 5, \quad y = \frac{25+20}{2} = 22.5$$Rashmi posts her flag on the 5th line at distance 22.5 m from AB (i.e., at point (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 ratio $k:1$.
Using x-coordinate:
$$-1 = \frac{6k + (-3)}{k+1} = \frac{6k-3}{k+1}$$$$-k - 1 = 6k - 3$$$$2 = 7k$$$$k = \frac{2}{7}$$So ratio = $k:1 = 2:7$.
Verification with y-coordinate:
$$y = \frac{-8k + 10}{k+1} = \frac{-8 \times \frac{2}{7} + 10}{\frac{2}{7}+1} = \frac{\frac{-16+70}{7}}{\frac{9}{7}} = \frac{54}{9} = 6 ✓$$∴ The point $(-1, 6)$ divides the segment in ratio 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
The x-axis has y = 0. Let the division point be $P(x, 0)$ dividing AB in ratio $k:1$.
$$0 = \frac{5k + (-5)}{k+1} = \frac{5k-5}{k+1}$$$$5k - 5 = 0 \Rightarrow k = 1$$So ratio = $1:1$ (midpoint).
$$x = \frac{-4 \times 1 + 1 \times 1}{1+1} = \frac{-4+1}{2} = \frac{-3}{2}$$∴ The x-axis divides AB in ratio 1 : 1 at point $\left(-\dfrac{3}{2}, 0\right)$.
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 $A(1,2)$, $B(4,y)$, $C(x,6)$, $D(3,5)$ be vertices of parallelogram ABCD.
Diagonals of parallelogram bisect each other, so midpoint of AC = 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:
$$\left(\frac{4+3}{2}, \frac{y+5}{2}\right) = \left(\frac{7}{2}, \frac{y+5}{2}\right)$$Equating:
$$\frac{1+x}{2} = \frac{7}{2} \Rightarrow x = 6$$$$4 = \frac{y+5}{2} \Rightarrow y+5 = 8 \Rightarrow y = 3$$∴ $x = 6$ and $y = 3$.
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
Centre is midpoint of diameter AB. Let $A(x, y)$.
$$\frac{x+1}{2} = 2 \Rightarrow x = 3$$$$\frac{y+4}{2} = -3 \Rightarrow y = -10$$∴ $A = \mathbf{(3, -10)}$.
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
$AP = \dfrac{3}{7}AB \Rightarrow AP : PB = 3 : 4$
(Since $PB = AB - AP = AB - \dfrac{3}{7}AB = \dfrac{4}{7}AB$)
Using section formula with $A(-2,-2)$, $B(2,-4)$, ratio $3:4$:
$$P_x = \frac{3 \times 2 + 4 \times (-2)}{3+4} = \frac{6-8}{7} = \frac{-2}{7}$$$$P_y = \frac{3 \times (-4) + 4 \times (-2)}{7} = \frac{-12-8}{7} = \frac{-20}{7}$$∴ $P = \left(-\dfrac{2}{7}, -\dfrac{20}{7}\right)$.
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 $P_1$, $P_2$, $P_3$ be the three points dividing AB into four equal parts.
$P_1$ (1:3 ratio):
$$P_1 = \left(\frac{1 \times 2 + 3 \times (-2)}{4}, \frac{1 \times 8 + 3 \times 2}{4}\right) = \left(\frac{2-6}{4}, \frac{8+6}{4}\right) = \left(-1, \frac{14}{4}\right) = \left(-1, \frac{7}{2}\right)$$$P_2$ = midpoint (1:1 ratio):
$$P_2 = \left(\frac{-2+2}{2}, \frac{2+8}{2}\right) = (0, 5)$$$P_3$ (3:1 ratio):
$$P_3 = \left(\frac{3 \times 2 + 1 \times (-2)}{4}, \frac{3 \times 8 + 1 \times 2}{4}\right) = \left(\frac{6-2}{4}, \frac{24+2}{4}\right) = \left(1, \frac{26}{4}\right) = \left(1, \frac{13}{2}\right)$$∴ The three points are $\left(-1, \dfrac{7}{2}\right)$, $(0, 5)$, $\left(1, \dfrac{13}{2}\right)$.
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 $A(3,0)$, $B(4,5)$, $C(-1,4)$, $D(-2,-1)$.
Diagonal AC:
$$d_1 = AC = \sqrt{(-1-3)^2 + (4-0)^2} = \sqrt{16+16} = 4\sqrt{2}$$Diagonal BD:
$$d_2 = BD = \sqrt{(-2-4)^2 + (-1-5)^2} = \sqrt{36+36} = 6\sqrt{2}$$Area of rhombus:
$$= \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = \frac{1}{2} \times 48 = 24 \text{ sq. units}$$∴ Area of rhombus = 24 sq. units.