Using determinants, find the area of the triangle with vertices (−3, 5), (3, −6), (7, 2).

Given:
Vertices of triangle: (− 3, 5), (3, − 6) and (7, 2)

Area of the triangle $=\Delta=\frac{1}{2}\left|\begin{array}{ccc}-3 & 5 & 1 \\ 3 & -6 & 1 \\ 7 & 2 & 1\end{array}\right|$

$=\frac{1}{2}\left|\begin{array}{ccc}-3 & 5 & 1 \\ 6 & -11 & 0 \\ 7 & 2 & 1\end{array}\right| \quad\left[\right.$ Applying $\left.R_{2} \rightarrow R_{2}-R_{1}\right]$

$=\frac{1}{2}\left|\begin{array}{ccc}-3 & 5 & 1 \\ 6 & -11 & 0 \\ 10 & -3 & 0\end{array}\right| \quad$ [Applying $R_{3} \rightarrow R_{3}-R_{1}$ ]

$=\frac{1}{2}\left|\begin{array}{cc}6 & -11 \\ 10 & -3\end{array}\right|$

$=\frac{1}{2}(-18+110)$

$=46$ square units