Ans: B
Let $(x,y)$ be the coordinates of $P$.
Let $(x,y)$ be the coordinates of $P$.
$\begin{array}{rcl}
AP & = & PB \\
\sqrt{(x-3)^2 + (y-9)^2} & = & \sqrt{(x-7)^2 + (y-1)^2} \\
x^2 -6x + 9 + y^2 -18y + 81 & = & x^2 -14x +49+y^2 -2y +1 \\
8x -16y +40 & = & 0 \\
2x – 4y + 10 & = & 0~\ldots \unicode{x2460}
\end{array}$
Sub. $y=x+1$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
2x -4(x+1) +10 & = & 0 \\
2x – 4x -4 + 10 & = & 0 \\
2x & = & 6 \\
x & = & 3
\end{array}$
Sub. $x=3$ into $y=x+1$, we have
$\begin{array}{rcl}
y & = & (3) + 1 \\
& = & 4
\end{array}$
Therefore, $P=(3,4)$.
