Question

If the point $A(2,-4)$ is equidistance from the points $P(3,8)$ and $Q(-10,y)$ find the value of $y$. Also find the value of $PQ$.

Answer 1

As we know that distance between the two points $(x_1,y_1)$ and $(x_2,y_2)$ is given as-

$d=\sqrt{{(x_2-{}_1)}^{[2}+{(y_2-y_1)}^2}$

Therefore,

$AP=\sqrt{{(3-2)}^2+{(8-(-4))}^2}=\sqrt{145}\text{units}$

$AQ=\sqrt{{(-10-2)}^2+{(y-(-4))}^2}=\sqrt{144+{(y+4)}^2}\text{units}$

$PQ=\sqrt{{(-10-3)}^2+{(y-8)}^2}=\sqrt{169+{(y-8)}^2}\text{units}$

Now,

$AP=AQ\left(\text{Given}\right)$

$\therefore \sqrt{145}=\sqrt{144+{(y+4)}^2}$

Squaring both sides, we have

$145=144+{(y+4)}^2$

$y^2+16+8y=1$

$\Rightarrow y^2+8y+15=0$

$\Rightarrow (y+5)(y+3)=0$

$\Rightarrow y=-5$ OR $y=-3$

Case I:- $y=-3$

$PQ=\sqrt{169+{(-3-8)}^2}=\sqrt{169+121}=\sqrt{290}\text{units}$

Case II:- $y=-5$

$PQ=\sqrt{169+{(-5-8)}^2}=\sqrt{169+169}=13\sqrt{2}\text{units}$