The area of the triangle whose vertices are (3, 8), (-4, 2) and (5, 1) is

$\frac{61}{2}$

The correct answer is Option (2) - $\frac{61}{2}$

Area = $\begin{vmatrix}\frac{1}{2}\begin{vmatrix}3&8&1\\-4&2&1\\5&1&1\end{vmatrix}\end{vmatrix}$

$R_2→R_2-R_3,R_1→R_1-R_3$

$=\begin{vmatrix}\frac{1}{2}\begin{vmatrix}-2&7&0\\-9&1&0\\5&1&1\end{vmatrix}\end{vmatrix}$

$=\left|\frac{1}{2}(-2+63)\right|$

$=\frac{61}{2}$ sq units