If the area of triangle with vertices A(1, 3), B(2, 0) and C(k, 2 k) is $\frac{k^2}{2}$ sq. units, then the value(s) of k is/are:

2 and 3

The correct answer is Option (3) - 2 and 3

area = $\frac{1}{2}\begin{Vmatrix}1&3&1\\2&0&1\\k&2k&1\end{Vmatrix}=\frac{k^2}{2}$

$R_2→R_2-R_1,R_3→R_3-R_1$

$\begin{Vmatrix}1&3&1\\1&-3&1\\k-1&2k-3&0\end{Vmatrix}=k^2$

$=2k-3+3k-3=k^2$

$k^2-5k+6=0$

$(k-2)(k-3)=0$

$⇒k=2,3$