Find the area of the triangle whose vertices are (2,7), (1,1) and (10,8) using determinants.

The correct answer is $\dfrac{47}{2}$ sq. units

Area of a triangle = $\triangle$ = $\dfrac{1}{2}$ $\left| \begin{matrix}

2 & \phantom{-}7 & \phantom{-}1 \\

1 & \phantom{-}1 & \phantom{-}1\\

10 & \phantom{-}8 & \phantom{-}1\\

\end{matrix} \right|$

$=$ $\dfrac{1}{2}$ $[2(1-8) - 7(1-10) +1 (8-10)]$
$=$ $\dfrac{47}{2}$ sq. units