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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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