Q19 of 50 Page 1

If the point P (k – 1, 2) is equidistant from the points A (3, k) and B (k, 5), find the values of k.

Given P (k – 1, 2) is equidistant from A (3, k) and B (k, 5).


Thus, AP = PB


We know that the distance between the points P (x1, y1) and Q (x2, y2) is


⇒ AP2 = PB2


⇒ (k – 1 – 3)2 + (2 – k)2 = (k – (k – 1))2 + (5 – 2)2


⇒ (k – 4)2 + (2 – k)2 = 12 + 32


⇒ k2 + 16 – 8k + 4 + k2 – 4k = 1 + 9


⇒ 2k2 – 12k + 20 = 10


⇒ 2k2 – 12k + 10 = 0


⇒ k2 – 6k + 5 = 0


⇒ k2 – k – 5k + 5 = 0


⇒ k (k – 1) – 5 (k – 1) = 0


⇒ (k – 1) (k – 5) = 0


⇒ k = 1, 5


∴ The values of k are 1 and 5.


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