# Chapter 8 - Section 8.5 - Determinants and Cramer's Rule - Exercise Set - Page 946: 50

The area of the triangle is $26$ square units.

#### Work Step by Step

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 given by: $\text{Area}=\pm \frac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|$ The area of a triangle whose vertices are $\left( 3,-5 \right),\left( 2,6 \right)\text{ and }\left( -3,5 \right)$ is given by: $\text{Area}=\pm \frac{1}{2}\left| \begin{matrix} 1 & 1 & 1 \\ -2 & -3 & 1 \\ 11 & -3 & 1 \\ \end{matrix} \right|$ Calculate the above determinant by expanding along the first row as follows: \begin{align} & \text{Area}=\frac{1}{2}\left[ 1\left| \begin{matrix} -3 & 1 \\ -3 & 1 \\ \end{matrix} \right|-1\left| \begin{matrix} -2 & 1 \\ 11 & 1 \\ \end{matrix} \right|+1\left| \begin{matrix} -2 & -3 \\ 11 & -3 \\ \end{matrix} \right| \right] \\ & =\frac{1}{2}\left[ 1\left( -3+3 \right)-1\left( -2-11 \right)+\left( 6+33 \right) \right] \\ & =\frac{1}{2}\left[ 1\left( 0 \right)-1\left( -13 \right)+39 \right] \\ & =\frac{1}{2}\left[ 13+39 \right] \end{align} Next, simplify it, \begin{align} & \text{Area}=\frac{1}{2}\left[ 13+39 \right] \\ & =\frac{1}{2}\times 52 \\ & =26 \end{align} Hence, $\text{Area}=26\text{ square units}$.

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