Find the distance between the points (2, 3) and (4, 1).

Using the distance formula, d = √[(4-2)² + (1-3)²] = √[2² + (-2)²] = √[4 + 4] = √8 = 2√2 units.

NCERT Solutions Class 10 Maths Chapter 7: Coordinate Geometry

Q: Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

The points are not collinear. By calculating the distances, AB = √5, BC = √212, and AC = √265. Since the sum of any two distances (AB + BC) is not equal to the third distance (AC), the points do not lie on a straight line.

Q: Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Yes, they are vertices of an isosceles triangle. Calculation shows AB = √37 and BC = √37. Since two sides are equal, it is an isosceles triangle.

Q: Find the point on the x-axis which is equidistant from (2, – 5) and (– 2, 9).

Let the point be (x, 0). Solving (x-2)² + (0 - (-5))² = (x - (-2))² + (0-9)² gives x = -7. Therefore, the point is (-7, 0).

Q: Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.

Using the distance formula, (10-2)² + (y + 3)² = 10². This simplifies to the quadratic equation y² + 6y - 27 = 0. Solving this gives y = 3 or y = -9.

Q: Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).

By setting the distances equal: (x-3)² + (y-6)² = (x+3)² + (y-4)². Expanding and simplifying leads to the relation 3x + y - 5 = 0.

Detailed solutions for Exercise 7.1 (Distance Formula).

Ex 7.1 Ex 7.2

#### **Q1. Find the distance between the following pairs of points:** **(i) (2, 3), (4, 1)** **(ii) (– 5, 7), (– 1, 3)** **(iii) (a, b), (– a, – b)** #### **A1. Solution:** **(i)** $x_1=2, y_1=3, x_2=4, y_2=1$ $d = \sqrt{(4-2)^2 + (1-3)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$ units. **(ii)** $x_1=-5, y_1=7, x_2=-1, y_2=3$ $d = \sqrt{(-1 - (-5))^2 + (3-7)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}$ units. **(iii)** $x_1=a, y_1=b, x_2=-a, y_2=-b$ $d = \sqrt{(-a-a)^2 + (-b-b)^2} = \sqrt{(-2a)^2 + (-2b)^2} = \sqrt{4a^2 + 4b^2} = 2\sqrt{a^2+b^2}$ units. #### **Q2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2?** #### **A2. Solution:** Points are (0, 0) and (36, 15). $d = \sqrt{(36-0)^2 + (15-0)^2} = \sqrt{36^2 + 15^2}$ $d = \sqrt{1296 + 225} = \sqrt{1521} = 39$. Yes, the distance between town A and B is 39 km. #### **Q3. Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.** #### **A3. Solution:** Let $A(1, 5), B(2, 3), C(-2, -11)$. $AB = \sqrt{(2-1)^2 + (3-5)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{5} \approx 2.24$ $BC = \sqrt{(-2-2)^2 + (-11-3)^2} = \sqrt{(-4)^2 + (-14)^2} = \sqrt{16 + 196} = \sqrt{212} \approx 14.56$ $AC = \sqrt{(-2-1)^2 + (-11-5)^2} = \sqrt{(-3)^2 + (-16)^2} = \sqrt{9 + 256} = \sqrt{265} \approx 16.28$ Since $AB + BC \neq AC$, the points are **not collinear**. #### **Q4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.** #### **A4. Solution:** Let $A(5, -2), B(6, 4), C(7, -2)$. $AB = \sqrt{(6-5)^2 + (4 - (-2))^2} = \sqrt{1^2 + 6^2} = \sqrt{37}$ $BC = \sqrt{(7-6)^2 + (-2-4)^2} = \sqrt{1^2 + (-6)^2} = \sqrt{37}$ $AC = \sqrt{(7-5)^2 + (-2 - (-2))^2} = \sqrt{2^2 + 0} = 2$ Since $AB = BC$, the triangle is **isosceles**. #### **Q5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.** #### **A5. Solution:** From the figure, coordinates are: $A(3, 4), B(6, 7), C(9, 4), D(6, 1)$. $AB = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$ $BC = \sqrt{3^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2}$ $CD = \sqrt{(-3)^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2}$ $DA = \sqrt{(-3)^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$ All sides are equal. Now check diagonals: $AC = \sqrt{(9-3)^2 + (4-4)^2} = \sqrt{6^2} = 6$ $BD = \sqrt{(6-6)^2 + (1-7)^2} = \sqrt{(-6)^2} = 6$ Since all sides are equal and diagonals are equal, it is a square. **Champa is correct.** #### **Q6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:** **(i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0)** **(ii) (– 3, 5), (3, 1), (0, 3), (– 1, – 4)** **(iii) (4, 5), (7, 6), (4, 3), (1, 2)** #### **A6. Solution:** **(i)** Let $A(-1, -2), B(1, 0), C(-1, 2), D(-3, 0)$. $AB = BC = CD = DA = \sqrt{8}$. Diagonals $AC = BD = 4$. Sides equal, diagonals equal $\Rightarrow$ **Square**. **(ii)** Let $A(-3, 5), B(3, 1), C(0, 3), D(-1, -4)$. Calculate $AB = \sqrt{52}, BC = \sqrt{13}, AC = \sqrt{13}$. Since $AC + BC = AB$ ($\sqrt{13} + \sqrt{13} = 2\sqrt{13} = \sqrt{52}$), points A, B, C are collinear. **No quadrilateral is formed.** **(iii)** Let $A(4, 5), B(7, 6), C(4, 3), D(1, 2)$. $AB = \sqrt{10}, BC = \sqrt{18}, CD = \sqrt{10}, DA = \sqrt{18}$. $AC = \sqrt{4} = 2, BD = \sqrt{52}$. Opposite sides equal, diagonals not equal $\Rightarrow$ **Parallelogram**. #### **Q7. Find the point on the x-axis which is equidistant from (2, – 5) and (– 2, 9).** #### **A7. Solution:** Let point on x-axis be $P(x, 0)$. Given $A(2, -5)$ and $B(-2, 9)$. $PA^2 = PB^2$ $(x-2)^2 + (0 - (-5))^2 = (x - (-2))^2 + (0-9)^2$ $x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81$ $-4x + 29 = 4x + 85$ $-8x = 56 \Rightarrow x = -7$. **Ans: (-7, 0)** #### **Q8. Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.** #### **A8. Solution:** $PQ = 10 \Rightarrow PQ^2 = 100$. $(10-2)^2 + (y - (-3))^2 = 100$ $8^2 + (y+3)^2 = 100$ $64 + y^2 + 6y + 9 = 100$ $y^2 + 6y - 27 = 0$ $(y+9)(y-3) = 0 \Rightarrow y = -9$ or $y = 3$. **Ans: y = 3, -9** #### **Q9. If Q(0, 1) is equidistant from P(5, – 3) and R(x, 6), find the values of x. Also find the distances QR and PR.** #### **A9. Solution:** $QP^2 = QR^2$ $(5-0)^2 + (-3-1)^2 = (x-0)^2 + (6-1)^2$ $25 + 16 = x^2 + 25$ $x^2 = 16 \Rightarrow x = \pm 4$. If $x = 4$: $QR = \sqrt{4^2 + 5^2} = \sqrt{41}$; $PR = \sqrt{(4-5)^2 + (6 - (-3))^2} = \sqrt{1+81} = \sqrt{82}$. If $x = -4$: $QR = \sqrt{(-4)^2 + 5^2} = \sqrt{41}$; $PR = \sqrt{(-4-5)^2 + (6 - (-3))^2} = \sqrt{81+81} = 9\sqrt{2}$. **Ans: $x = \pm 4, QR = \sqrt{41}, PR = \sqrt{82} \text{ or } 9\sqrt{2}$** #### **Q10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).** #### **A10. Solution:** Let $P(x, y), A(3, 6), B(-3, 4)$. $PA^2 = PB^2$ $(x-3)^2 + (y-6)^2 = (x+3)^2 + (y-4)^2$ $x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16$ $-6x - 12y + 45 = 6x - 8y + 25$ $12x + 4y = 20$ $3x + y = 5$ (Dividing by 4) **Ans: $3x + y - 5 = 0$**