Ans: (a) $A’=(3,4)$, $B’=(5,-2)$ (b) $x-3y-1=0$
- $A’=(3,4)$, $B’=(5,-2)$.
- Note that the locus of $P$ is the perpendicular bisector of $A’B’$.
The mid-point of $A’B’$
$\begin{array}{cl}
= & \left(\dfrac{3+5}{2}, \dfrac{4+(-2)}{2}\right) \\= & (4,1)
\end{array}$The slope of $A’B’$
$\begin{array}{cl}
= & \dfrac{-2-4}{5-3} \\
= & -3
\end{array}$The slope of the locus of $P$
$\begin{array}{cl}
= & -1 \div -3 \\
= & \dfrac{1}{3}
\end{array}$Therefore, the equation of the locus of $P$ is
$\begin{array}{rcl}
y-1 & = & \dfrac{1}{3}(x-4) \\
x – 3y – 1 & = & 0
\end{array}$
