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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.
Solution
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.
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