# RD Sharma Class 10 Ex 14.2 Solutions Chapter 14 Coordinate Geometry

In this chapter, we provide RD Sharma Class 10 Ex 14.2 Solutions Chapter 14 Coordinate Geometry for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 14.2 Solutions Chapter 14 Coordinate Geometry pdf, Now you will get step by step solution to each question.

# Chapter 14: Coordinate Geometry Exercise – 14.2

### Question: 1

Find the distance between the following pair of points:

(i) (- 6, 7) and (-1, -5)

(ii) (a + b, b + c) and (a -b, c – b)

(iii) (a sin α, – b cos α) and (- a cos α, b sin α)

(iv) (a, 0) and (0, b)

### Solution:

(i) We have P (- 6, 7) and Q (- 1, – 5)

Here,

x1 = – 6, y1 = 7 and

x2 = -1, Y2 = – 5

(ii) We have P (a + b, b + c) and Q (a – b, c – b) here,

x1 = a + b, y= b + c and x2 = a – b, Y2 = c – b

(iii) We have P(a sinα, – b cos α) and Q(-a cos α, b sin α) here

x= a sin α, y1 = – b cos α and

x– a cos α, y2 = b sin α

(iv) We have P(a, 0) and Q (0, b)

Here,

x1 = a,y1 = 0, x2 = 0, y2 = b,

### Question: 2

Find the value of a when the distance between the points (3, a) and (4, 1) is √10.

### Solution:

We have P (3, a) and Q(4, 1)

Here,

Squaring both sides

⇒ 10 = 2 + a2 – 2a

⇒ a2 – 2a + 2 – 10 = 0

⇒ a2 – 2a – 8 = 0

Splitting the middle team

⇒ a2 – 4a + 2a – 8 = 0

⇒ a(a – 4) + 2(a – 4) = 0

⇒ (a – 4) (a + 2) = 0

⇒ a = 4, a = – 2

### Question: 3

If the points (2, 1) and (1, -2) are equidistant from the point (x, y) from (-3, 0) as well as from (3, 0) are 4.

### Solution:

We have P(2, 1) and Q(1,- 2) and R(X, Y)

Also, PR = QR

∴ PR = QR

⇒ x+ 5 – 4x + y2 –2y = x+ 5 – 2x + y+ 4y

⇒ x+ 5 – 4x + y2 – 2y = x+ 5 – 2x + y+ 4y

⇒ – 4x + 2x – 2y – 4y = 0

⇒ – 2x – 6y = 0

⇒ – 2(x + 3y) = 0

⇒ -2(x + 3y) = 0

⇒ x + 3y = 0/-2

⇒ x + 3y = 0

Hence Proved.

### Question: 4

Find the value of x, y if the distances of the point (x, y) from (- 3, 0) as well as from (3, 0) are 4.

### Solution:

We have P(x, y), Q( -3, 0) and R(3, 0)

Squaring both sides

⇒ 16 = x2 + 9 + 6x + y2

⇒ x2 + y2 = 7 – 6x                            …… (1)

Squaring both sides

⇒16 = x2 + 9 – 6x + y2

⇒ x2 + y2 = 16 – 9 + 6x

⇒ x2+ y= 7 + 6x                   …. (2)

Equating (1) and (2)

7 – 6x = 7 + 6x

⇒ 7 – 7 = 6x + 6x

⇒ 0 = 12x

⇒ x = 12

Substituting the value of x = 0 in (2)

x2 + y2 = 7+ 6x

0 + y2 = 7 + 6 × 0

Y2 = 7

Y = + 7

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