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If Q($0$, $1$) is equidistant from P($5$, $-3$) and R($x$, $6$), find the values of $x$.
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PQ = QR $\Rightarrow PQ^2 = QR^2$
$(5-0)^2 + (-3-1)^2 = (x - 0)^2 + (6 - 1)^2$
$\Rightarrow 25 + 16 = x^2 + 25$
$\Rightarrow x^2 = 16$
$\Rightarrow x = 4, x = -4$
$(5-0)^2 + (-3-1)^2 = (x - 0)^2 + (6 - 1)^2$
$\Rightarrow 25 + 16 = x^2 + 25$
$\Rightarrow x^2 = 16$
$\Rightarrow x = 4, x = -4$