Find the point on the straight line $3 x + y + 4 = 0$ which is equidistant from the points $(- 5, 6)$ and $(3, 2)$ .

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Solution

Let the point be $(x, y)$ as it lies on $3 x + y + 4 = 0$
$\Rightarrow y = - 3 x - 4$
$\Rightarrow$ Point $= (x, - 3 x - 4)$
Distance of this is equidistant from $(- 5, 6)$ and $(3, 2)$
$\Rightarrow \sqrt{(x + 5)^{2} + (6 + 3 x + 4)^{2}} = \sqrt{(x - 3)^{2} + (2 + 3 x + 4)^{2}}$
$\Rightarrow (x + 5)^{2} + (3 x + 10)^{2} = (x - 3)^{2} + (3 x + 6)^{2}$
$\Rightarrow x^{2} + 25 + 10 x + 9 x^{2} + 100 + 60 x = x^{2} + 9 - 6 x + 9 x^{2} + 36 - 36 x$
$\Rightarrow 70 x + 125 = 30 x + 45$
$\Rightarrow 40 x = - 80$
$\Rightarrow x = - 2$
$\Rightarrow y = - 3 (- 2) - 4 = 2$
$\Rightarrow$ Point $= (- 2, 2)$ .

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