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If the point P (k -1, 2) is equidistant from the points A (3, k) and B (k,5), find the value of k. |
Answer» Coordinates of points are A(3, k), B(k, 5) and P(k-1, 2) Using distance formula = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) ⇒ PA = PB \(\sqrt{(k - 1-3)^2 + (2 - k)^2}\) = \(\sqrt{(k - 1-k)^2 + (2 - 5)^2}\) On squaring both sides, we get (k - 1 - 3)2 + (2 - k)2 = (k - 1 - k)2 + (2 - 5)2 (k - 4)2 + (2 - k)2 = (-1)2 + (-3)2 k2 - 8k + 16 + k2 - 4k + 4 = 1 + 9 k2 - 6k + 5 = 0 k2 - 5k - k + 5 = 0 k(k - 5) - 1(k - 5) = 0 (k - 5)(k - 1) = 0 k = 1, 5 |
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