Question

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.

Answer 1

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". `