# If the point P(2, 2) is equidistant from the points A(−2, k) and B(−2k, −3), find k. Also find the length of AP. - Mathematics

If the point P(2, 2) is equidistant from the points A(−2, k) and B(−2k, −3), find k. Also find the length of AP.

#### Solution

The given points are P(2, 2), A(−2, k) and B(−2k, −3).

We know that the distance between the points,(x1,y1) and (x2,y2)is given by:

d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)

It is given that P is equidistant from A and B.
∴ AP = BP
⇒ AP2 = BP2
⇒ (2 − (−2))2 + (2 − k)2 = (2 − (−2k))2 + (2 − (−3))2
⇒ (4)2 + (2 − k)2 = (2 + 2k)2 + (5)2
⇒ 16 + k2 + 4 − 4k = 4 + 4k2 + 8k + 25
⇒ 3k2 + 12k + 9 = 0
⇒ k2 + 4k + 3 = 0
⇒ k2 + 3k + k + 3 = 0
⇒ (k + 1) (k + 3) = 0
⇒ k = −1, −3

Thus, the value of k is −1 and −3.

For k = −1:
Length of AP = sqrt((2-(-2))^2+(2-1(-1))^2)=sqrt(4^2+3^2)=sqrt(16+9)=sqrt25=5

For k = −3:
Length of AP =sqrt((2-(-2))^2+(2-1(-3))^2)=sqrt(4^2+5^2)=sqrt(16+25)=sqrt41

Thus, the length of AP is either 5 " units" or sqrt41 "units".

Concept: Distance Formula
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Exercise 6.2 | Q 49 | Page 17