Question

Find the coordinates of the point equidistant from the given points.
A(5, 1), B(-3, -7) and C(7, -1).
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Answer 1

Answer

Let the required point be $\sf P (x, y)$.

Then,

$\sf \dashrightarrow PA = PB = PC$

$\sf \dashrightarrow {PA}^{2} = {PB}^{2} = {PC}^{2}$

$\sf \dashrightarrow {PA}^{2} = {PB}^{2} \: \: and \: \: {PB}^{2} = {PC}^{2}$

$\sf \dashrightarrow \begin{cases} \sf {(x - 5)}^{2} + {(y - 1)}^{2} = {(x + 3)}^{2} + {(y + 7}^{2} \\ \sf {(x + 3)}^{2} + {(y + 7)}^{2} = {(x - 7)}^{2} + {(y + 1)}^{2}\end{cases}$

$\sf \dashrightarrow \begin{cases} \sf {x}^{2} + {y}^{2} - 10x - 2y+ 26 = {x}^{2} + {y}^{2} + 6x + 14y + 58 \\ \sf {x}^{2} + {y}^{2} + 6x + 14y + 58 = {x}^{2} + {y}^{2} - 14x + 2y + 50\end{cases}$

$\sf \dashrightarrow {x}^{2} + {y}^{2} - 10x - 2y + 26 = {x}^{2} + {y}^{2} + 6x + 14y + 58$

$\sf \dashrightarrow 16x + 16y = -32$

$\sf \dashrightarrow x + y = -2 \: \: --- (i)$

$\sf \dashrightarrow {x}^{2} + {y}^{2} - 6x + 14y + 58 = {x}^{2} + {y}^{2} - 14x + 2y + 50$

$\sf \dashrightarrow 20x + 12y = -8$

$\sf \dashrightarrow 5x + 3y = -2 \: \: --- (ii)$

Multiply (i) by 5 and subtract (ii) from the result.

$\sf \dashrightarrow 2y = -8$

$\sf \dashrightarrow y = -4$

Putting value of y in equation (i), we get x = 2.

So,

$\sf \dashrightarrow x = 2 \: \: and \: \: y = -4$

Thus, the required point is $\sf P(2, -4)$.