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

#### Solution

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

We know that the distance between the points,(x_{1},y_{1}) and (x_{2},y_{2})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

⇒ AP^{2} = BP^{2}

⇒ (2 − (−2))^{2} + (2 − k)^{2} = (2 − (−2k))^{2} + (2 − (−3))^{2}

⇒ (4)^{2} + (2 − k)^{2} = (2 + 2k)^{2} + (5)^{2}

⇒ 16 + k^{2} + 4 − 4k = 4 + 4k^{2} + 8k + 25

⇒ 3k^{2} + 12k + 9 = 0

⇒ k^{2} + 4k + 3 = 0

⇒ k^{2} + 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". `