CBSE Class 10 Chapter 7 Coordinate Geometry Exercise 7.1

Q1.	Find the distance between the following pairs of points :
(i)	(2, 3), (4, 1)
Ans.	Given: $$(x_{1,},y_{1})=(2, 3)$$ $$(x_{2,},y_{2})=(4, 1)$$ We know that distance formula is: $$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$$ Putting value: $$d=\sqrt{\left(4-2\right)^{2}+\left(1-3\right)^{2}}$$ $$d=\sqrt{\left(2\right)^{2}+\left(-2\right)^{2}}$$ $$d=\sqrt{4+4}$$ $$d=\sqrt{8}$$ $$d=2\sqrt{2}$$

(ii)

(—5, 7), (—1, 3)

Ans.

Given: $$(x_{1,},y_{1})=(—5, 7)$$ $$(x_{2,},y_{2})=(—1, 3)$$ We know that distance formula is: $$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$$ Putting value: $$d=\sqrt{\left((—1)-(—5)\right)^{2}+\left(3-7\right)^{2}}$$ $$d=\sqrt{\left((—1)+5\right)^{2}+\left(3-7\right)^{2}}$$ $$d=\sqrt{\left(4\right)^{2}+\left(-4\right)^{2}}$$ $$d=\sqrt{16+16}$$ $$d=\sqrt{32}$$ $$d=4\sqrt{2}$$

(iii)	(a, b), (—a, —b)
Ans.	Given: $$(x_{1,},y_{1})=(a, b)$$ $$(x_{2,},y_{2})=(—a, —b)$$ We know that distance formula is: $$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$$ Putting value: $$d=\sqrt{\left(-a-a\right)^{2}+\left(-b-b\right)^{2}}$$ $$d=\sqrt{\left(-2a\right)^{2}+\left(-2b\right)^{2}}$$ $$d=\sqrt{4a^{2}+4b^{2}}$$ $$d=2\sqrt{a^{2}+b^{2}}$$

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.

Ans.

Given: $$(x_{1,},y_{1})=(0, 0)$$ $$(x_{2,},y_{2})=(36, 15)$$ We know that distance formula is: $$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$$ Putting value: $$d=\sqrt{\left(36-0\right)^{2}+\left(15-0\right)^{2}}$$ $$d=\sqrt{\left(36\right)^{2}+\left(15\right)^{2}}$$ $$d=\sqrt{1296+225}$$ $$d=\sqrt{1521}$$ $$d=39$$ So, the distance between the points (0, 0) and (36, 15) is 39 units. In section 7.2, The distance between two town is: A is (4, 0) and B is (6, 0) $$d=\sqrt{\left(6-4\right)^{2}+\left(0-0\right)^{2}}$$ $$d=\sqrt{\left(2\right)^{2}+\left(0\right)^{2}}$$ $$d=\sqrt{4+0}$$ $$d=\sqrt{4}$$ $$d=2$$ So, the distance between two town point A (0, 0) and B (36, 15) is 2 units.

Q3.

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

Ans.

Given: We know that If the sum of the lengths of any two line segments equals the length of the third line segment, then the three points are collinear. Let A (1,5), B(2,3) and C(– 2, – 11) then Find the distance between points; say AB, BC and CA and assume AB = d1, BC = d2, CA = d3 We also know that distance formula is: $$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$$ Putting value for d1: $$d_{1}=\sqrt{\left(2-1\right)^{2}+\left(3-5\right)^{2}}$$ $$d_{1}=\sqrt{\left(1\right)^{2}+\left(-2\right)^{2}}$$ $$d_{1}=\sqrt{1+4}$$ $$d_{1}=\sqrt{5}$$ Putting value for d2: $$d_{2}=\sqrt{\left((-2)-2\right)^{2}+\left((-11)-3\right)^{2}}$$ $$d_{2}=\sqrt{\left(-4\right)^{2}+\left(-14\right)^{2}}$$ $$d_{2}=\sqrt{16+196}$$ $$d_{2}=\sqrt{212}$$ Putting value for d3: $$d_{3}=\sqrt{\left((-2)-1\right)^{2}+\left((-11)-5\right)^{2}}$$ $$d_{3}=\sqrt{\left(-3\right)^{2}+\left(-16\right)^{2}}$$ $$d_{3}=\sqrt{9+256}$$ $$d_{3}=\sqrt{265}$$ Since, $$d_{1}+d_{2}\neq d_{3}$$ $$AB+BC\neq CA$$ So, the points (1, 5), (2, 3) and (– 2, – 11) are not collinear. Other Method: If the two slopes are equal, the three points are collinear. To determine if the points (1, 5), (2, 3), and (–2, –11) are collinear, we can check if the slopes between consecutive pairs of points are equal or not. We know that (m) between two points (x1, y1) and (x2, y2) given by $$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$ Slope between (1, 5) and (2, 3): $$m_{1}=\frac{3-5}{2-1}=\frac{-2}{1}=-2$$ Slope between (2, 3) and (–2, –11): $$m_{2}=\frac{-11-3}{-2-2}=\frac{14}{4}=\frac{7}{2}$$ Since, $$m_{1}\neq m_{2}$$ So, the points (1, 5), (2, 3) and (– 2, – 11) are not collinear.

Q4.

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

Ans.

We know that In Isosceles triangle, any two sides are equal. So, we need to check the distances between pairs of points. Given:  (5, – 2), (6, 4) and (7, – 2) are vertices. We also know that distance formula is: $$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$$ Distance between (5, –2) and (6, 4): $$d_{1}=\sqrt{\left(6-5)\right)^{2}+\left(4-(-2)\right)^{2}}=\sqrt{\left(1\right)^{2}+\left(6\right)^{2}}=\sqrt{1+36}=\sqrt{37}$$ Distance between (6, 4) and (7, –2): $$d_{2}=\sqrt{\left(7-6\right)^{2}+\left((-2)-4\right)^{2}}=\sqrt{\left(1\right)^{2}+\left(6\right)^{2}}=\sqrt{1+36}=\sqrt{37}$$ Distance between (7, –2) and (5, –2): $$d_{3}=\sqrt{\left(5-7\right)^{2}+\left((-2)-(-2)\right)^{2}}=\sqrt{\left(-2\right)^{2}+\left(0\right)^{2}}=\sqrt{4+0}=\sqrt{4}=2$$ Since, $$d_{1}=d_{2}=\sqrt{37}$$ Therefore, the points (5, –2), (6, 4), and (7, –2) are the vertices of an isosceles triangle.

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.

Ans.

If all four sides are equal in length and both diagonals are equal, then ABCD is a square. If any of these conditions is not met, then it is not a square. Let's the coordinates of A, B, C and D is  $$A(x_{1},y_{1})=(3,4),B(x_{2},y_{2})=(6,7), C(x_{3},y_{3})=(9,4), D(x_{4},y_{4})=(6,1)$$ The distances between consecutive pairs of points: $$AB=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$$ $$AB=\sqrt{(6-3)^{2}+(7-4)^{2}}=\sqrt{3^{2}+3^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$$ $$BC=\sqrt{(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}$$ $$BC=\sqrt{(9-6)^{2}+(4-7)^{2}}=\sqrt{3^{2}+(-3)^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$$ $$CD=\sqrt{(x_{4}-x_{3})^{2}+(y_{4}-y_{3})^{2}}$$ $$CD=\sqrt{(6-9)^{2}+(1-4)^{2}}=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$$ $$DA=\sqrt{(x_{1}-x_{4})^{2}+(y_{1}-y_{4})^{2}}$$ $$DA=\sqrt{(3-6)^{2}+(4-1)^{2}}=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}3\sqrt{2}$$ Calculate the diagonals: $$AC=\sqrt{(x_{3}-x_{1})^{2}+(y_{3}-y_{1})^{2}}$$ $$AC=\sqrt{(9-3)^{2}+(4-4)^{2}}=\sqrt{6^{2}+0^{2}}=\sqrt{36+0}=\sqrt{36}=6$$ $$BD=\sqrt{(x_{4}-x_{2})^{2}+(y_{4}-y_{2})^{2}}$$ $$BD=\sqrt{(6-6)^{2}+(1-7)^{2}}=\sqrt{0^{2}+(-6)^{2}}=\sqrt{0+36}=\sqrt{36}=6$$ Since, $$AB = BC =CD = DA$$ $$AC = BD$$ Therefore, ABCD is a square.

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)
Ans.	Here, A(– 1, – 2), B(1, 0), C(– 1, 2), D(– 3, 0) So by distance formula $$AB=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$$ $$AB=\sqrt{(1-(-1))^{2}+(0-2)^{2}}=\sqrt{2^{2}+2^{2}}=\sqrt{4+4}=\sqrt{8}$$ $$BC=\sqrt{(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}$$ $$BC=\sqrt{(-1-1)^{2}+(2-0)^{2}}=\sqrt{2^{2}+(2)^{2}}=\sqrt{4+4}=\sqrt{8}$$ $$CD=\sqrt{(x_{4}-x_{3})^{2}+(y_{4}-y_{3})^{2}}$$ $$CD=\sqrt{((-3)-(-1))^{2}+(0-2)^{2}}=\sqrt{(-2)^{2}+(2)^{2}}=\sqrt{4+4}=\sqrt{8}$$ $$DA=\sqrt{(x_{1}-x_{4})^{2}+(y_{1}-y_{4})^{2}}$$ $$DA=\sqrt{(-1-3)^{2}+(-2-0)^{2}}=\sqrt{(2)^{2}+(-2)^{2}}=\sqrt{4+4}=\sqrt{8}$$ $$AC=\sqrt{(x_{3}-x_{1})^{2}+(y_{3}-y_{1})^{2}}$$ $$AC=\sqrt{(-1-(-1))^{2}+(2-(-2))^{2}}=\sqrt{0^{2}+4^{2}}=\sqrt{0+16}=\sqrt{16}=4$$ $$BD=\sqrt{(x_{4}-x_{2})^{2}+(y_{4}-y_{2})^{2}}$$ $$BD=\sqrt{(-3-1)^{2}+(0-0)^{2}}=\sqrt{4^{2}+0^{2}}=\sqrt{16+0}=\sqrt{16}=4$$ Since, $$AB = BC =CD = DA$$ $$AC = BD$$ Therefore, Quadrilateral formed by (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0) is a square.

(ii)

(–3, 5), (3, 1), (0, 3), (–1, – 4)

Ans.

Here, A(–3, 5), B(3, 1), C(0, 3), D(–1, – 4) So by distance formula $$AB= \sqrt{(-3 - 3)^2 + (5 - 1)^2} = \sqrt{36 + 16} = \sqrt{52}$$ $$BC= \sqrt{(0 - 3)^2 + (3 - 1)^2} = \sqrt{9 + 4} = \sqrt{13}$$ $$CD= \sqrt{(-1 - 0)^2 + ((-4) - 3)^2} = \sqrt{1 + 49} = \sqrt{50}$$ $$DA= \sqrt{(-3 - (-1))^2 + (5 - (-4))^2} = \sqrt{4 + 81} = \sqrt{85}$$ The distances are: $$AB = \sqrt{52}, \quad BC = \sqrt{13}, \quad CD = \sqrt{50}, \quad DA = \sqrt{85}$$ Since all four sides have different lengths. so, we cannot determine a specific type like square, rectangle, rhombus, or kite based on these distances alone.

(iii)

(4, 5), (7, 6), (4, 3), (1, 2)

Ans.

Here, A(4, 5), B(7, 6), C(4, 3), D(1, 2) So by distance formula $$AB= \sqrt{(7 - 4)^2 + (6 - 5)^2} = \sqrt{9 + 1} = \sqrt{10}$$ $$BC= \sqrt{(4 - 7)^2 + (3 - 6)^2} = \sqrt{9 + 9} = \sqrt{18}$$ $$CD= \sqrt{(1 - 4)^2 + (2 - 3)^2} = \sqrt{9 + 1} = \sqrt{10}$$ $$DA= \sqrt{(4 - 1)^2 + (5 - 2)^2} = \sqrt{9 + 9} = \sqrt{18}$$ $$AC= \sqrt{(4 - 4)^2 + (3 - 5)^2} = \sqrt{0 +4} = 2$$ $$BD= \sqrt{(1 - 7)^2 + (2 - 6)^2} = \sqrt{36 +16} = 13\sqrt{2}$$ The distances are: $$AB = \sqrt{10}, \quad BC = \sqrt{18}, \quad CD = \sqrt{10}, \quad DA = \sqrt{18}$$ $$AC\neq BD$$ Since opposite sides have equal lengths (AB = CD and BC = DA) and diagonal have different lengths, the quadrilateral is a parallelogram.

Q7.

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

Ans.

Let the point on the x-axis be $(x, 0)$. We have to find point on x-axis. So, y-coordinate will be 0. The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ in a plane is given by the distance formula: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ In this case, the distance between $(x, 0)$ and $(2, -5)$ is equal to the distance between $(x, 0)$ and $(-2, 9)$. So, we can write as: $$\sqrt{(x - 2)^2 + (0 - (-5))^2} = \sqrt{(x - (-2))^2 + (0 - 9)^2}$$ Simplify this equation and solve for $x$: $$\sqrt{(x - 2)^2 + 25} = \sqrt{(x + 2)^2 + 81}$$ $$(x - 2)^2 + 25 = (x + 2)^2 + 81$$ $$x^2 - 4x + 29 = x^2 + 4x + 85$$ $$\text{Subtract } x^2 \text{ from both sides:}$$ $$-4x + 29 = 4x + 85$$ $$\text{Subtract } 4x \text{ from both sides:}$$ $$29 = 8x + 85$$ $$\text{Subtract } 85 \text{ from both sides:}$$ $$-56 = 8x$$ $$\text{Divide by } 8 \text{ to isolate } x:$$ $$x = -7$$ So, the point on the x-axis that is equidistant from $(2, -5)$ and $(-2, 9)$ is $(-7, 0)$.

Q8.

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

Ans.

We have to find the values of $y$ for which the distance between the points $P(2, -3)$ and $Q(10, y)$ is 10 units. By using distance formula: $$10 = \sqrt{(10 - 2)^2 + (y - (-3))^2}$$ Solving for $y$: $$10 = \sqrt{8^2 + (y + 3)^2}$$ $$10 = \sqrt{64 + (y + 3)^2}$$ $$10^2 = 64 + (y + 3)^2$$ $$100 = 64 + (y + 3)^2$$ $$(y + 3)^2 = 100 - 64$$ $$(y + 3)^2 = 36$$ Taking the square root of both sides: $$y + 3 = \pm 6$$ Solving for $y$: 1. $y + 3 = 6$: $y = 3$ 2. $y + 3 = -6$: $y = -9$ Therefore, the values of $y$ for which the distance between $P(2, -3)$ and $Q(10, y)$ is 10 units are $y = 3$ and $y = -9$.

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.

Ans.

Given that $Q(0,1)$ is equidistant from $P(5,-3)$ and $R(x,6)$, we can use the distance formula. The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by: $$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ $Q(0,1)$ is equidistant from $P(5,-3)$ and $R(x,6)$, So $$\sqrt{(5-0)^2 + (-3-1)^2} = \sqrt{(x-0)^2 + (6-1)^2}$$ $$\sqrt{25 + 16} = \sqrt{x^2 + 25}$$ Squaring both sides: $$41 = x^2 + 25$$ $$x^2 = 16$$ $$x = \pm 4$$ Therefore, point $R$ is $(4,6)$ or $(-4,6)$. Case (1): When $R$ is $(4,6)$, $$\text{Distance PR} = \sqrt{(5-4)^2 + (-3-6)^2} = \sqrt{1 + 81} = \sqrt{82}$$ $$\text{Distance QR} = \sqrt{(0-4)^2 + (1-6)^2} = \sqrt{16 + 25} = \sqrt{41}$$ Case (2): When $R$ is $(-4,6)$, $$\text{Distance PR} = \sqrt{(5-(-4))^2 + (-3-6)^2} = \sqrt{81 + 81} = 9\sqrt{2}$$ $$\text{Distance QR} = \sqrt{(0-(-4))^2 + (1-6)^2} = \sqrt{16 + 25} = \sqrt{41}$$

Q10.

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

Ans.

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ In this case, the point $(x, y)$ is equidistant from $(3, 6)$ and $(-3, 4)$, so the distances $d_1$ and $d_2$ are equal: $$\sqrt{(x - 3)^2 + (y - 6)^2} = \sqrt{(x - (-3))^2 + (y - 4)^2}$$ Squaring both sides to eliminate the square roots: $$(x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2$$ $$x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16$$ $$-6x - 12y + 45 = 6x - 8y + 25$$ $$-6x -6x - 12y + 8y =  - 20$$ $$-12x + 4y = - 20$$ Divide both side by 4: $$-3x + y = - 5$$ So, the relation between $x$ and $y$ is $-3x + y = -5$, or alternatively, $y = 3x - 5$. This equation represents the set of points $(x, y)$ that are equidistant from $(3, 6)$ and $(-3, 4)$.