If area of the triangle with vertices A(2, k), B(1, 1) and C(10, 8) is $\frac{47}{2}$ sq. units, then one of the values of k is

7

The correct answer is Option (4) - 7

area = $\frac{1}{2}\begin{Vmatrix}2&k&1\\1&1&1\\10&8&1\end{Vmatrix}=\frac{47}{2}$

$R_1→R_1-R_2,R_3→R_3-R_2$

$\begin{Vmatrix}1&k-1&0\\1&1&1\\9&7&0\end{Vmatrix}=47$

$=9k-9-7=47$

$9k=63⇒k=7$