1.

Find the value of k, if the point P(0, 2) is equidistant from (3, k) and (k, 5).

Answer»

Let the point P (0, 2) is equidistant from A (3, k) and B (k, 5)

So, PA = PB

PA2 = PB2

(3 -0)2 + (k -2)2 = (k – 0)2 + (5 – 2)2

9 + k2 + 4 – 4k – k2 – 9 = 0

4 – 4k = 0

-4k = -4

Therefore, the value of k = 1



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