Find the coordinates of the point equidistant from three given points A(5, 3), B(5, -5) and C(1, -5).

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Solution

Let the required point be P(x,y). Then $A P = B P = C P$

That is, $(A P)^{2} = (B P)^{2} = (C P)^{2}$

This means, $(A P)^{2} = (B P)^{2}$

$\Rightarrow\Rightarrow (x - 5)^{2} + (y - 3)^{2} = (x - 5)^{2} + (y + 5)^{2}$

$\Rightarrow\Rightarrow x^{2} - 10 x + 25 + y^{2} - 6 y + 9 = x^{2} - 10 x + 25 + y 2 + 10 y + 25$

$\Rightarrow\Rightarrow x^{2} + y^{2} - 10 x - 6 y + 34 = x^{2} + y^{2} - 10 x + 10 y + 50$

$\Rightarrow\Rightarrow x^{2} + y^{2} - 10 x - 6 y - x^{2} - y^{2} + 10 x - 10 y = 50 - 34$

$\Rightarrow\Rightarrow - 16 y = 16$

$\Rightarrow\Rightarrow y = - 16 / 16 = - 1$

And $(B P)^{2} = (C P)^{2}$

$\Rightarrow\Rightarrow (x - 5)^{2} + (y + 5)^{2} = (x - 1)^{2} + (y + 5)^{2}$

$\Rightarrow\Rightarrow x^{2} - 10 x + 25 + y^{2} + 10 y + 25 = x^{2} - 2 x + 1 + y^{2} + 10 y + 25$

$\Rightarrow\Rightarrow x^{2} + y^{2} - 10 x + 10 y + 50 = x^{2} + y^{2} - 2 x + 10 y + 26$

$\Rightarrow\Rightarrow x^{2} + y^{2} - 10 x + 10 y - x^{2} - y^{2} + 2 x - 10 y = 26 - 50$

$\Rightarrow\Rightarrow - 8 x = - 24$

$\Rightarrow\Rightarrow y = - 24 / - 8 = 3$

Hence the required point is (3,-1)

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