Find the co-ordinates of the point equidistant from the points $A (- 2, - 3), B (- 1, 0)$ and $C (7, - 6)$

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Solution

Given,
the point is equidistant from $A (- 2, - 3) B (- 1, 0)$ and $C (7, - 6)$
Let the point be $(x, y) = D$
$A D = B D = C D$
$\sqrt{(x + 2)^{2} + (y + 3)^{2}} = \sqrt{(x + 1)^{2} + (y - 6)^{2}} = \sqrt{(x - 7)^{2} + (y + 6)^{2}}$
Squaring on all sides
$\Rightarrow (x + 2)^{2} + (y + 3)^{2} = (x + 1)^{2} + y^{2} = (x - 7)^{2} + (y + 6)^{2}$
$\Rightarrow x^{2} + 4 x + 4 + y^{2} + 6 y + 9 y = x^{2} + 2 x + 1 + y^{2} = x^{2} - 14 x + 49 + y^{2} + 12 y + 36$
$\Rightarrow 4 x - 6 y + 13 = 2 x + 1 = - 14 x + 12 y + 85$
Taking first and second equation
$\Rightarrow 4 x + 6 y + 13 = 2 x + 1$
$\Rightarrow 2 x + 6 y = - 12$ , $x + 3 y = - 6$ ……(i)
Taking first and third equation
$\Rightarrow 4 x + 6 y + 13 = - 14 x + 12 y + 85$
$\Rightarrow 18 x - 6 y = 72$ , $3 x - y = 12$ ……(ii)
By equation (i) and (ii)
$3 x + 9 y = - 18$
$3 x - y = 12$
We get $8 y = - 6$
$\Rightarrow y = - \frac{3}{4}$ and $x \Rightarrow \frac{15}{4}$
$∴ (x, y) = (\frac{15}{4}, \frac{- 3}{4})$ .

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