Ans: D
Note that the three perpendicular bisectors of the three sides of the triangle pass through the circumcentre. The mid-point of the two vertices
Note that the three perpendicular bisectors of the three sides of the triangle pass through the circumcentre. The mid-point of the two vertices
$\begin{array}{cl}
= & \left( \dfrac{-4+6}{2}, \dfrac{-8+2}{2} \right) \\
= & ( 1, -3)
\end{array}$
Therefore, we have
$\begin{array}{rcl}
\dfrac{-8 – 2}{-4 – 6} \times \dfrac{-4 – (-3)}{k-1} & = & -1 \\
\dfrac{-1}{k-1} & = & -1 \\
-1 & = & -(k-1) \\
-1 & = & -k +1 \\
k & = & 2
\end{array}$
