Find the coordinates of the point equidistant from the three 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 AP = BP = CP
That is, (AP)²=(BP)²=(CP)²
This means (AP)²=(BP)²
$\Rightarrow {(x - 5)}^{2} + {(y - 3)}^{2} = {(x - 5)}^{2} + {(y + 5)}^{2} \Rightarrow x^{2} - 10 x + 25 + y^{2} - 6 y + 9 = x^{2} - 10 x + 25 + y^{2} + 10 y + 25 \Rightarrow x^{2} - 10 x + y^{2} - 6 y + 34 = x^{2} - 10 x + y^{2} + 10 y + 50 \Rightarrow x^{2} - 10 x + y^{2} - 6 y - x^{2} + 10 x - y^{2} - 10 y = 50 - 34 \Rightarrow - 16 y = 16 \Rightarrow y = - \frac{16}{16} = - 1 And {(B P)}^{2} = {(C P)}^{2} \Rightarrow {(x - 5)}^{2} + {(y + 5)}^{2} = {(x - 1)}^{2} + {(y + 5)}^{2} \Rightarrow x^{2} - 10 x + 25 + y^{2} + 10 y + 25 = x^{2} - 2 x + 1 + y^{2} + 10 y + 25 \Rightarrow x^{2} - 10 x + y^{2} + 10 y + 50 = x^{2} - 2 x + y^{2} + 10 y + 26 \Rightarrow x^{2} - 10 x + y^{2} + 10 y - x^{2} + 2 x - y^{2} - 10 y = 26 - 50 \Rightarrow - 8 x = - 24 \Rightarrow x = \frac{- 24}{- 8} = 3$
Hence, the required point is (3, −1).

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