Well-designed AP 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Exercise 7.1 offers step-by-step explanations to help students understand problem-solving strategies.

## Coordinate Geometry Class 10 Exercise 7.1 Solutions – 10th Class Maths 7.1 Exercise Solutions

Question 1.

Find the distance between the following pairs of points:

i) (2, 3), (4, 1)

ii) (- 5, 7), (- 1, 3)

iii) (a, b), (- a, – b)

Solution:

i) Let A(2, 3) and B(4, 1)

(x_{1} y_{1}) (x_{2} y_{2})

are the points on line AB.

We know that distance between two

points is \(\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}\)

AB distance = \(\sqrt{(4-2)^2+(1-3)^2}\)

Therefore, AB = \(\sqrt{2^2+(-2)^2}\)

= \(\sqrt{4+4}\)

= \(\sqrt{8}\) or 2\(\sqrt{2}\) units.

ii) Let P(-5, 7) and Q(-1, 3)

(x_{1} y_{1}) (x_{2} y_{2})

are the points on line PQ.

We know that distance between two

points is \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

PQ distance = \(\sqrt{(-1-(-5))^2+(3-7)^2}\)

= \(\sqrt{(-1+5)^2+(-4)^2}\)

= \(\sqrt{4^2+16}\)

Therefore PQ = \(\sqrt{16+16}\)

= \(\sqrt{32}\) = 4\(\sqrt{2}\) units

iii) Let K(a, b) and L(-a, -b)

(x_{1} y_{1}) (x_{2} y_{2})

are the points on line KL.

We know that distance between two

points is \(\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}\)

distance = \(\sqrt{(-a-a)^2+(-b-b)^2}\)

= \(\sqrt{(-2 a)^2+(-2 b)^2}\)

Therefore, KL = \(\sqrt{4 a^2+4 b^2}\)

= \(\sqrt{4\left(a^2+b^2\right)}\)

= \(2 \sqrt{a^2+b^2}\) units

Question 2.

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.

Solution:

Given A(0, 0) and B(36, 15) are two towns.

Distance between them is AB.

In ∆ABC, ∠C = 90°

By Pythagoras theorem,

AB^{2} = AC^{2} + BC^{2} = 36^{2} + 15^{2}

AB = \(\sqrt{1296+225}\) = 39 km.

Therefore, distance between two towns AB = 39 km.

(Or)

Distance between A(0, 0) and B(36, 15) is AB.

x_{1} = 0, y_{1} = 0; x_{2} = 36, y_{2} = 15

We know that distance between two

points = \(\sqrt{\left(x_2-x_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}\)

= \(\sqrt{(36-0)^2+(15-0)^2}\)

= \(\sqrt{36^2+15^2}\)

= \(\sqrt{1296+225}\)

AB = \(\sqrt{1521}\) = 39 km.

∴ Distance between two towns A and B is 39 km.

Question 3.

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

Solution:

Let the points are A(1, 5), B(2, 3) and C(-2, -11).

Distance between two points

AB + BC ≠ AC

So, A, B, C are not collinear points.

Question 4.

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

Solution:

Let A(5, -2), B(6, 4) and C(7, -2) are the vertices of ∆ABC.

Distance between two points

Clearly we got AB = BC

So, ∆ABC is an isosceles triangle.

Question 5.

In a classroom, 4 friends are seated at the points A, B, C and D as shown in given figure. 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.

Solution:

From the given figure, points are A(3, 4), B(6, 7), C(9, 4) and D(6, 1).

Distance between two points

Therefore, AB = BC = CD = AD = 3\(\sqrt{2}\) units

Diagonals AC = BD = 6 units

So, all sides and two diagonals are equal.

Therefore, given vertices can form a square.

So, ABCD is a square.

Clearly, Champa said ABCD is a square is correct.

Question 6.

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)

Solution:

i) Let A(-1, -2), B(1, 0), C(-1, 2) and D(-3, 0) are vertices of a quadrilateral.

In this quadrilateral AB = BC = CD = AD = 2\(\sqrt{2}\) units and diagonals AC = BD = 4 units.

So, given points can form a square,

ii) Let P(-3, 5), Q(3, 1), R(0, 3), S(-1, -4) are vertices of quadrilateral PQRS.

Distance between two points

So, all sides of the quadrilateral are of different lengths.

Hence, given points cannot form a special quadrilateral.

iii) Let A(4, 5), D(7, 6), I(4, 3), N(1, 2) are the vertices of the quadrilateral ADIN.

Distance between two points

In the quadrilateral opposite sides are equal and diagonals are not equal. Hence given point can form a parallelogram.

Question 7.

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

Solution:

Let the P(x, 0) is the point on the X-axis which is equidistant from the points A(2, -5) and B(-2, 9), that is AP = PB.

Distance between two points

Question 8.

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

Solution:

Given points are P(2, -3) and Q(10, y) and PQ = 10 units.

Distance between two points

⇒ 64 + y^{2} + 6y + 9 = 100

⇒ y^{2} + 6y + 73 – 100 = 0

⇒ y^{2} + 6y – 27 = 0

⇒ y^{2} + 9y – 3y – 27 = 0

⇒ y(y + 9) – 3(y + 9) = 0

⇒ (y – 3)(y + 9) = o

Therefore y = 3 (or) -9

Question 9.

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.

Solution:

Q(0, 1) is equidistant from P(5, -3) and R(x, 6) that is QR = PQ

Distance between two points

Question 10.

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

Solution:

Let P(x, y) is equidistant from the points A(3, 6) and B(-3, 4) that is PB = PA

Distance between two points

-6x – 6x – 12y + 8y = 25 – 45

-12x – 4y = -20

12x + 4y = 20

⇒ 3x + y = 5