Question

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

• A. $(-2,-4)$
• B. $(2,-4)$
• C. $(2,4)$
• D. $(-2,4)$

Answer 1

The correct option is B $(2,-4)$

Let the required point be $P(p,q).$

Then the distances $PA=PB=PC=d$ as $A(5,1),B(-3,-7)$ & $C(7,-1)$ are equidistant from $P.$

Applying the distance formula,

$P(p,q)\equiv (x_1,y_1)$ and $A(5,1)\equiv (x_2,y_2)$

$d=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2},$

$d_{PA}=\sqrt{{(p-5)}^2+{(q-1)}^2},$

$P(p,q)\equiv (x_1,y_1)$ and $B(-3,-7)\equiv (x_2,y_2)$

$d_{PB}=\sqrt{{(p+3)}^2+{(q+7)}^2}$ and

$P(p,q)\equiv (x_1,y_1)$ and $C(7,-1)\equiv (x_2,y_2)$

$d_{Pc}=\sqrt{{(p-7)}^2+{(q+1)}^2}$

$\therefore \sqrt{{(p-5)}^2+{(q-1)}^2}=\sqrt{{(p+3)}^2+{(q+7)}^2}$

$⟹-16p-16q=32$

$⟹p+q=-2$ ........(i)

$\sqrt{{(p+3)}^2+{(q+7)}^2}=\sqrt{{(p-7)}^2+{(q+1)}^2}$

$⟹20p+12q=-8$

$⟹5p+3q=-2$ .........(ii).

Multiplying (i) by 5 & subtracting from (ii),

$-2q=8$

$⟹q=-4.$

Substituting $q=-4$ in (i),

$p=-2-q=-2+4=2.$

$\therefore P(p,q)=P(2,-4).$