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

As per the question, we have

AP = BP

`⇒sqrt((2+2)^2 +(2+k)^2) = sqrt(( 2+2k)^2 +(2+3)^2)`

`⇒sqrt((4)^2 +(2-k)^2) = sqrt((2+2k)^2 + (5)^2)`

⇒ 16+ 4 +k² - 4k = 4+ 4k² + 8k +25        (Squaring both sides) 

`⇒k^2 + 4k +3=0`

⇒ (k+1) (k+3) =0

⇒ k =-3, -1

Now for k = -1

`AP= sqrt ((2+2)^2 +(2-k)^2)`

`= sqrt((4)^2 +(2+1)^2)`

`= sqrt(16+9) = 5` units 

For k = -3 

`AP= sqrt(( 2+2)^2 +(2-k)^2)`

`= sqrt((4)^2+(2+3)^2)`

`=sqrt(16+25) = sqrt(41)` units

Hence, k= -1,-3; AP= 5 units for k=-1 and AP=`sqrt(41)` units for k=-3.