Students can practice TS Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 to get the best methods of solving problems.

## TS 10th Class Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.1

Question 1.

Find the distance between the following pair of points.

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

Solution:

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

Solution:

iii) (-2, -3) and (3, 2)

Solution:

iv) (a, b) and (-a, -b)

Solution:

Question 2.

Find the distance between the points (0, 0) and (36, 15)

Solution:

Given : origin O(0, 0) and a point P(36, 15)

Distance between any point and origin = \(\sqrt{x^2+y^2}\)

Question 3.

Verify whether the points (1, 5),(2, 3) and (-2, -1) are colleniar or not.

Solution:

Given : A(1, 5) B(2, 3) and C(-2, -1)

Here the sum of no two line segments is equal to third segment

Hence the points are not collinear.

slope of AB, m_{1} = \(\frac{3-5}{2-1}\) = -2

slope of BC, m_{2} = \(\frac{-1-3}{-2-2}\) = -2

m_{1} ≠ m_{2}

Hence A, B, C are not collinear.

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)

Now we have AB = BC.

∴ Δ ABC is an isosceles triangle.

i.e., given points are the vertices of an isosceles triangle.

Question 5.

In a classroom, 4 friends are seated at the points A, B, C and D as shown in figure. Jarina and Phani walk into the class and after observing for a few minutes Jarina asks Phani ‘Don’t you think ABCD ¡s a square ?“ Phani disagrees. Using distance formula. find which of them is correct. Why?

Solution:

Given : Four friends are seated at A, B, C and D where A(3, 4), B(6, 7), C(9, 4) and D(6, 1)

Hence in ABCD four sides are equal.

i.e., AB = BC = CD = DA = 3\(\sqrt{2}\) units.and two diagonals are equal.

ie., AC = BD = 6 units.

∴ ABCD forms a square

i.e., Jarina is correct.

Question 6.

Show that the following points form an equilateral triangle A(a, 0), B(-a, 0), C(0, a\(\sqrt{3}\)).

Solution:

Now AB = BC = CA.

∴ Δ ABC is an equilateral triangle.

Question 7.

Prove that the points (-7, -3), (5, 10), (15, 8) and (3, -5) taken in order are the corners of a parallelogram and find its area.

(A.P. Mar. ’15)

Solution:

Diagram

Given : A(-7, -3), B(5, 10), C(15, 8) and D(3, -5)

In ABCD, both pairs of opposite sides (AB, CD) and (BC, AD) are equal.

Hence the given points form a parallelogram.

Area of ABCD = 2 × ΔABC

Question 8.

Show that the points (-4, -7), (-1, 2), (8, 5) and (5, -4) taken in order are the vertices of a rhombus.

Hint : Area of rhombus = \(\frac{1}{2}\) × product (A.P. June ‘is’)

Solution:

Given in ABCD, A (-4, -7), B(-1, 2), C(8, 5) and D(5, -4)

Distance formula = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

∴ In ABCD, AB = BC = CD = AD [from sides are equal]

Hence ABCD is a rhombus.

Area of a rhombus = \(\frac{1}{2}\)d_{1}d_{2}

= \(\frac{1}{2}\) × 12\(\sqrt{2}\) × 6\(\sqrt{2}\)

= 72 sq. units.

Question 9.

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)

Solution:

Let A(-1, -2), B(1, 0), C(-1, 2), D(-3, 0) be the given points.

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

= \(\sqrt{16}\) = 4 units.

BD = \(\sqrt{(-3-1)^2+(0-0)^2}\)

= \(\sqrt{16}\) = 4 units.

In ABCD, AB = BC = CD = AD four sides are equal.

AC = BD → diagonals are equal.

Hence, the given points form a square.

ii) (-3, 5), (3, 1), (1, -3), (-1, -4)

Solution:

Let A(-3, 5), B(3, 1), C(1, -3) and D(-1, -4) be the given points.

∴ Opposite sides are not equal.

∴ Its diagonal are not equal.

In ABCD, AB ≠ CD, BC ≠ AD and AC ≠ BD

Hence ABCD is not a parallelogram

∴ The given points can’t be form a parallelogram.

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

Solution:

Let A(4, 5), B(7, 6), C(4, 3) and D(1, 2) be the given points.

In ABCD, AB = CD and BC = AD (i.e., both pairs of opposite sides are equal) and AC ≠ BD

Hence ABCD is a parallelogram, i.e., the given points form a parallelogram.

Question 10.

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

Solution:

Given points, A(2, -5), B(-2, 9)

Let P(x, 0) be the point on x-axis.

Which is equidistant from A and B i.e., PA = PB.

But PA = PB

⇒ \(\sqrt{x^2-4 x+29}\) = \(\sqrt{x^2+4 x+85}\)

squaring on both sides, we get

x^{2} – 4x + 29 = x^{2} + 4x + 85

-4x – 4x = 85 – 29

-8x = 56

∴ (x, 0) = (-8, 0) is the point which is equidis¬tant from the given points.

Question 11.

If the distance between two points (x, 7) and (1, 15) is 10, find the value of x.

Solution:

Given = A(x, 7), B(1, 15) and AB = 10.

Distance formula

squaring on both sides, we get.

\(\left[\sqrt{x^2-2 x+65}\right]^2\) = 10^{2}

⇒ x^{2} – 2x + 65 = 100

x^{2} – 2x – 35 = 0

⇒ x^{2} – 7x + 5x – 35 = 0

x(x- 7) + 5(x – 7) = 0

(x – 7) (x + 5) = 0

(x – 7) = 0 or x + 5 = 0

x = 7 or x = -5

∴ x = 7 or – 5

Question 12.

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

Solution:

Given : P(2, -3), Q(10, y) and \(\overline{\mathrm{PQ}}\) = 10

Distance formula

squaring on both sides we get,

\(\left[\sqrt{y^2+6 y+73}\right]^2\) = 10^{2}

⇒ y^{2} + 6y + 73 = 100

⇒ y^{2} + 9y – 3y – 27 = 0

⇒ y(y + 9) – 3 (y + 9) = 0

⇒ (y + 9) (y – 3) = 0

⇒ y + 9 = 0 or y – 3 = 0

⇒ y = – 9 or y = 3

⇒ y = – 9 or 3

Question 13.

Find the radius of the circle whose centre is (3, 2) and passes through (-5, 6).

Solution:

Given : A circle with centre A(3, 2) passing through B(-5, 6).

Radius = AB

[∵ Distance of a point from the centre of the circle.]

Distance formula

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

Radius r = \(\sqrt{(-5-3)^2+(6-2)^2}\)

= \(\sqrt{64+16}\) = \(\sqrt{80}\) = 4\(\sqrt{5}\) units.

Question 14.

Can you draw a triangle with vertices (1, 5), (5, 8) and (13, 14) ? Give reason.

Solution:

Let A(1, 5), B(5, 8) and C(13, 14) be the given points.

Distance formula

Here, AC = AB + BC

∴ Δ ABC can’t be formed with the given vertices.

[∵ sum of the any two sides of a triangle must be greater than the third side],

Question 15.

Find a relation between x and y such that the point (x, y) is equidistant from the points (-2, 8) and (-3, -5).

Solution:

Let A(-2, 8), B(-3, -5) and P(x, y)

If P is equidistant from A, B then PA = PB

Distance formula

Now PA = PB

⇒ \(\sqrt{x^2+y^2+4 x-16 y+68}\)

⇒ \(\sqrt{x^2+y^2+6 x+10 y+34}\)

squaring on both sides we get,

⇒ 4x – 16y – 6x – 10y = 34 – 68

⇒ – 2x – 26y = – 34

⇒ x + 13y = 17 is the required condition.