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.

AP = BP

AP = √{(– 2 – 2)2 + (k – 2)2}


= √{16 + k2 – 4k + 4}


= √(k2 – 2k + 20)


BP = √{(– 2k – 2)2 + (– 3 – 2)2}


= √{4k2 + 8k + 4 + 25}


= √(4k2 + 8k + 29)


Squaring AP and BP and equating them we get


k2 – 4k + 20 = 4k2 + 8k + 29


3k2 + 12k + 9 = 0


(k + 3)(k + 1) = 0


⇒ k = – 3


⇒ AP = √41units


Or k = – 1


⇒ AP = 5 units


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