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If the point P $(3, - 3)$ is equidistant from the points A $(4, 9)$ and B $(- 9, k)$, find the value(s) of $k$.
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$(3-4)^2 + (-3 - 9)^2 = (-9 - 3)^2 + (k - (-3))^2$
$(-1)^2 + (-12)^2 = (-12)^2 + (k+3)^2$
$1 + 144 = 144 + (k+3)^2$
$1 = (k+3)^2$
$k+3 = \pm 1$
$k = -2, -4$
$(-1)^2 + (-12)^2 = (-12)^2 + (k+3)^2$
$1 + 144 = 144 + (k+3)^2$
$1 = (k+3)^2$
$k+3 = \pm 1$
$k = -2, -4$