Question

If $A(x,3)$ lies on perpendicular bisector of the line segment joining the point $P(4,8)$ and $Q(-4,0)$ find $x$.

Answer 1

Given points are $P(4,8)$ and $Q(-4,0)$.

Let midpoint of $P$ and $Q$ be $D$.

Hence, the coordinates of the midpoint of $PQ$,

$D=(\frac{4-4}{2},\frac{8+0}{2})=(0,4)$.

Let the Slope of the line $PQ$ be $m=\frac{8-0}{4+4}=1$.

Since slope of line $PQ$ is $1$, any line perpendicular to $PQ$ will have slope $\frac{-1}{1}=-1$

Now consider line $AD$ , slope of line $AD=\frac{-1}{x}$

But slope of $PD=-1$ since it is perpendicular to $PQ$

Thus, $\frac{-1}{x}=-1$

$\therefore x=1$