Question

Find the coordinates of the point equidistant from the points $A(-2,-3)$, $B(-1,0)$ and $C(7,-6)$

• A. (3, -3)
• B. (-3,3)
• C. (-2,-3)
• D. (-3,-3)

Answer 1

The correct option is A (3, -3)

Given-

Points $A(x_1,y_1),B(x_2,y_2)$ & $C(x_3,y_3)$ are equidistant from a point P.

To find out-

The coordinate of P

Solution-

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

Then, by the given condition $PA=PB=PC.$

We shall use distance formula $d=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$

So, $PA=\sqrt{{(x_1-x)}^2+{(y_1-y)}^2}=\sqrt{{(2+x)}^2+{(3+y)}^2}$

$PB=\sqrt{{(x_2-x)}^2+{(y_2-y)}^2}=\sqrt{{(1+x)}^2+{(0+y)}^2}$

$PC=\sqrt{{(x_3-x)}^2+{(y_3-y)}^2}=\sqrt{{(7-x)}^2+{(6+y)}^2}$

$\therefore PA=PB$

$⟹\sqrt{{(2+x)}^2+{(3+y)}^2}=\sqrt{{(1+x)}^2+{(0+y)}^2}$

$⟹2x+6y+12=0$ ..........(i).

Again, $PB=PC$

$⟹\sqrt{{(1+x)}^2+{(0+y)}^2}=\sqrt{{(7-x)}^2+{(6+y)}^2}$

$⟹8x-6y-42=0$ ...........(ii).

Multiplying (i) by 4 and subtracting from (ii),

$30y=90$

$⟹y=-3.$

Substituting $y=3$ in (i),

$2x+6\times (-3)+12=0$

$⟹x=3.$

$\therefore$ The required point is $P(3,-3).$