College Algebra, Chapter 7, 7.4, Section 7.4, Problem 58

Use a determinant to find the area of the triangle with vertices $(-2,5), (7,2), (3, -4)$ and sketch the triangle.








$
\begin{equation}
\begin{aligned}

\text{area } = \pm \frac{1}{2} \left| \begin{array}{ccc}
-2 & 5 & 1 \\
7 & 2 & 1 \\
3 & -4 & 1
\end{array} \right| =& \pm \frac{1}{2} \left[ -2 \left| \begin{array}{cc}
2 & 1 \\
-4 & 1
\end{array} \right| - 5 \left| \begin{array}{cc}
7 & 1 \\
3 & 1
\end{array} \right| + 1 \left| \begin{array}{cc}
7 & 2 \\
3 & -4
\end{array} \right| \right]
\\
\\
=& \pm \frac{1}{2} \left[ -2 (2 \cdot 1 - 1 \cdot (-4)) - 5 (7 \cdot 3 - 1 \cdot 3) + (7 \cdot (-4) - 2 \cdot 3) \right]
\\
\\
=& \frac{-1}{2} (-66) \quad \text{to make the area positive, we use negative sign in the formula}
\\
\\
=& 33 \text{ square units}

\end{aligned}
\end{equation}
$