Ans: B
Let $C=(h,k)$. Since $C$ is lying on $x-2y=0$, we have
Let $C=(h,k)$. Since $C$ is lying on $x-2y=0$, we have
$\begin{array}{rcl}
h – 2k & = & 0 \\
k & = & \dfrac{h}{2} \ \ldots \unicode{x2460}
\end{array}$
Since $AC=BC$, we have
$\begin{array}{rcl}
\sqrt{(h-9)^2+(k-(-2))^2} & = & \sqrt{(h-(-1))^2+(k-8)^2} \\
h^2 -18h + 81 + k^2 +4k + 4 & = & h^2 +2h+1+k^2-16k+64 \\
-20h +20k +20 & = & 0 \\
h-k-1 & = & 0 \ \ldots \unicode{x2461}
\end{array}$
Sub. $\unicode{x2460}$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
h – \dfrac{h}{2} -1 & = & 0 \\
2h – h – 2 & = & 0 \\
h & = & 2
\end{array}$
