NCERT Solutions Class 10 Maths Chapter 7: Coordinate Geometry

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

Using the section formula, the coordinates are calculated as x = [2(4) + 3(-1)] / (2+3) = 1 and y = [2(-3) + 3(7)] / (2+3) = 3. The coordinates of the point are (1, 3).

Q: 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 section formula for the x-coordinate: -1 = (6k - 3) / (k + 1). Solving this gives -k - 1 = 6k - 3, which results in 7k = 2 or k = 2/7. The ratio is 2 : 7.

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

Since diagonals of a parallelogram bisect each other, the midpoint of (1, 2) and (x, 6) is the same as the midpoint of (4, y) and (3, 5). Equating the midpoints, we get (1+x)/2 = 7/2 which gives x = 6, and (2+6)/2 = (y+5)/2 which gives y = 3.

Q: 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).

The centre (2, -3) is the midpoint of diameter AB. Let A be (x, y). Then (x+1)/2 = 2 implies x = 3, and (y+4)/2 = -3 implies y = -10. The coordinates of A are (3, -10).

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

Area of a rhombus = 1/2 × (product of diagonals). Diagonal 1 (AC) = √32 = 4√2 and Diagonal 2 (BD) = √72 = 6√2. Area = 1/2 × 4√2 × 6√2 = 24 square units.

Detailed solutions for Exercise 7.2 (Section Formula).

Ex 7.1 Ex 7.2

#### **Q1. Find the coordinates of the point which divides the join of (– 1, 7) and (4, – 3) in the ratio 2 : 3.** #### **A1. Solution:** Let $P(x, y)$ be the required point. $x_1 = -1, y_1 = 7$ $x_2 = 4, y_2 = -3$ $m_1 = 2, m_2 = 3$ $x = \frac{2(4) + 3(-1)}{2+3} = \frac{8 - 3}{5} = \frac{5}{5} = 1$ $y = \frac{2(-3) + 3(7)}{2+3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3$ **Ans: The coordinates are (1, 3).** #### **Q2. Find the coordinates of the points of trisection of the line segment joining (4, – 1) and (– 2, – 3).** #### **A2. Solution:** Trisection means dividing into 3 equal parts. Let points be $P$ and $Q$. $P$ divides $AB$ in ratio $1 : 2$. $x = \frac{1(-2) + 2(4)}{1+2} = \frac{6}{3} = 2$; $y = \frac{1(-3) + 2(-1)}{1+2} = \frac{-5}{3}$. $Q$ divides $AB$ in ratio $2 : 1$. $x = \frac{2(-2) + 1(4)}{2+1} = \frac{0}{3} = 0$; $y = \frac{2(-3) + 1(-1)}{2+1} = \frac{-7}{3}$. **Ans: Coordinates are $(2, -\frac{5}{3})$ and $(0, -\frac{7}{3})$.** #### **Q3. 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. 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 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?** #### **A3. Solution:** Total distance AD = 100 m. Niharika's position (Green flag): $x = 2$, $y = \frac{1}{4} \times 100 = 25$. So, $G(2, 25)$. Preet's position (Red flag): $x = 8$, $y = \frac{1}{5} \times 100 = 20$. So, $R(8, 20)$. Distance $GR = \sqrt{(8-2)^2 + (20-25)^2} = \sqrt{6^2 + (-5)^2} = \sqrt{36 + 25} = \sqrt{61}$ m. Blue flag (Rashmi) is at midpoint of GR: $x = \frac{2+8}{2} = 5$; $y = \frac{25+20}{2} = 22.5$. **Ans: Distance is $\sqrt{61}$ m. Rashmi should post her flag on the 5th line at 22.5 m.** #### **Q4. Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).** #### **A4. Solution:** Let the ratio be $k : 1$. Using section formula for $x$-coordinate: $-1 = \frac{k(6) + 1(-3)}{k+1}$ $-k - 1 = 6k - 3$ $2 = 7k \Rightarrow k = \frac{2}{7}$. **Ans: The ratio is 2 : 7.** #### **Q5. 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.** #### **A5. Solution:** Point on x-axis is $P(x, 0)$. Let ratio be $k : 1$. Using $y$-coordinate: $0 = \frac{k(5) + 1(-5)}{k+1} \Rightarrow 5k - 5 = 0 \Rightarrow k = 1$. Ratio is $1 : 1$. Now find $x$: $x = \frac{1(-4) + 1(1)}{1+1} = \frac{-3}{2}$. **Ans: Ratio is 1 : 1. Point is $(-\frac{3}{2}, 0)$.** #### **Q6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.** #### **A6. Solution:** In a parallelogram, diagonals bisect each other. So, midpoint of diagonal AC = midpoint of diagonal BD. Midpoint of AC: $(\frac{1+x}{2}, \frac{2+6}{2}) = (\frac{1+x}{2}, 4)$ Midpoint of BD: $(\frac{4+3}{2}, \frac{y+5}{2}) = (\frac{7}{2}, \frac{y+5}{2})$ Equating coordinates: $\frac{1+x}{2} = \frac{7}{2} \Rightarrow 1+x = 7 \Rightarrow x = 6$ $4 = \frac{y+5}{2} \Rightarrow 8 = y+5 \Rightarrow y = 3$ **Ans: x = 6, y = 3.** #### **Q7. 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).** #### **A7. Solution:** Let $A$ be $(x, y)$. Centre $(2, -3)$ is the midpoint of AB. $2 = \frac{x+1}{2} \Rightarrow 4 = x+1 \Rightarrow x = 3$ $-3 = \frac{y+4}{2} \Rightarrow -6 = y+4 \Rightarrow y = -10$ **Ans: A is (3, -10).** #### **Q8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that $AP = \frac{3}{7} AB$ and P lies on the line segment AB.** #### **A8. Solution:** $AP = \frac{3}{7} AB \Rightarrow PB = AB - \frac{3}{7} AB = \frac{4}{7} AB$. Ratio $AP : PB = 3 : 4$. $x = \frac{3(2) + 4(-2)}{3+4} = \frac{6-8}{7} = -\frac{2}{7}$ $y = \frac{3(-4) + 4(-2)}{3+4} = \frac{-12-8}{7} = -\frac{20}{7}$ **Ans: P is $(-\frac{2}{7}, -\frac{20}{7})$.** #### **Q9. Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.** #### **A9. Solution:** Let points be $P_1, P_2, P_3$. $P_2$ is midpoint of AB: $x = \frac{-2+2}{2} = 0, y = \frac{2+8}{2} = 5$. So $P_2(0, 5)$. $P_1$ is midpoint of $AP_2$: $x = \frac{-2+0}{2} = -1, y = \frac{2+5}{2} = 3.5$. So $P_1(-1, 3.5)$. $P_3$ is midpoint of $P_2B$: $x = \frac{0+2}{2} = 1, y = \frac{5+8}{2} = 6.5$. So $P_3(1, 6.5)$. **Ans: (-1, 3.5), (0, 5), and (1, 6.5).** #### **Q10. 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 = $\frac{1}{2}$ (product of its diagonals))** #### **A10. Solution:** Let $A(3, 0), B(4, 5), C(-1, 4), D(-2, -1)$. Diagonal $d_1 (AC) = \sqrt{(-1-3)^2 + (4-0)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}$. Diagonal $d_2 (BD) = \sqrt{(-2-4)^2 + (-1-5)^2} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36+36} = \sqrt{72} = 6\sqrt{2}$. Area = $\frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = \frac{1}{2} \times 24 \times 2 = 24$ sq units. **Ans: 24 square units.**