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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

x1 = a sin α, y1 = - b cos α and

x2 – 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

⇒ x2 + 5 - 4x + y2 –2y = x2 + 5 - 2x + y2 + 4y

⇒ x2 + 5 - 4x + y2 – 2y = x2 + 5 - 2x + y2 + 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+ y2 = 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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