## Precalculus (6th Edition) Blitzer

$\text{Area}=28\text{ square units}$
The area of a triangle with vertices $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\text{ and }\left( {{x}_{3}},{{y}_{3}} \right)$ is: $\text{Area}=\pm \frac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|$ The area of a triangle whose vertices are $\left( 3,-5 \right),\left( 2,6 \right)\text{ and }\left( -3,5 \right)$ is: $\text{Area}=\pm \frac{1}{2}\left| \begin{matrix} 3 & -5 & 1 \\ 2 & 6 & 1 \\ -3 & 5 & 1 \\ \end{matrix} \right|$ Calculate the above determinant by expanding along the first row as follows: \begin{align} & \text{Area}=\frac{1}{2}\left[ 3\left| \begin{matrix} 6 & 1 \\ 5 & 1 \\ \end{matrix} \right|+5\left| \begin{matrix} 2 & 1 \\ -3 & 1 \\ \end{matrix} \right|+1\left| \begin{matrix} 2 & 6 \\ -3 & 5 \\ \end{matrix} \right| \right] \\ & =\frac{1}{2}\left[ 3\left( 6-5 \right)+5\left( 2+3 \right)+\left( 10+18 \right) \right] \\ & =\frac{1}{2}\left[ 3\left( 1 \right)+5\left( 5 \right)+\left( 28 \right) \right] \\ & =\frac{1}{2}\left[ 3+25+28 \right] \end{align} Next, we simplify it, \begin{align} & \text{Area}=\frac{1}{2}\left[ 3+25+28 \right] \\ & =\frac{1}{2}\times 56 \\ & =28 \end{align} Therefore, $\text{Area}=28\text{ square units}$.