Question

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

Answer 1

$\bf \underline{SOLUTION}:-$

$\bf \: •Let \: the \: required \: point \: be \: P(x,y).Then,$

$\bf PA=PB=PC$

$\bf \implies PA²=PB²=PC²$

$\bf \implies PA²=PB² \: and \: PB²=PC²$

$\implies \begin{cases}</p><p></p><p> \bf(x - 5) {}^{2} + (y - 1) {}^{2} = (x + 3) {}^{2} + (y + 7) {}^{2} \\</p><p></p><p> \bf( x + 3) {}^{2} + (y + 7) {}^{2} = (x - 7) {}^{2} + (y + 1) { }^{2} </p><p></p><p>\end{cases}</p><p>$

$\implies\begin{cases} \bf</p><p></p><p>x {}^{2} + y {}^{2} - 10x - 2y + 26 = x {}^{2} + y {}^{2} + 6x + 14y + 58 \\ \bf \: x {}^{2} </p><p></p><p> + y {}^{2} + 6x + 14y + 58 = x {}^{2} + y {}^{2} - 14x + 2y + 5 0\end{cases}</p><p>$

$\bf \: •Now \: x²+y²-10x-2y+26=x²+y²+6x+14y+58$

$\implies \bf16 x + 16y = - 32$

$\implies \bf \: x + y = - 2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: ....(1)$

$\bf •And \: x²+y²+6x+14y+58=x²+y²-14x+2y+50$

$\implies \bf \: 20x + 12y = - 8$

$\bf \implies \: 5x + 3y = - 2 \: \: \: \: \: \: \: \: ....(2)$

$\bf•Multiplying (1) \: by \: 5 ,we \: get$

$\bf •5x+5y=-10........(3)$

$\bf •Subtracting (2) \: from \: (3) \: we \: get$

$\bf \implies \cancel{5x} + 5y - \cancel{5x} - 3y = - 10 + 2$

$\implies \bf 2y = - 8$

$\bf \implies y = \frac { - \cancel{8} \: 4}{ \cancel{2}}$

$\boxed{ \bf\therefore \: y = - 4 }$

$\bf •Now \: putting \: the \: value \: of \underline{y=-4} \: in \: (1)$

$\implies \bf \: x + ( - 4) = - 2$

$\boxed{ \bf \therefore \: x = 2}$

Hence,the required point is P(2,-4)

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