Chapter 8.5, Problem 22E

To determine

To find:

Thearea of a triangle with the given vertices

Answer to Problem 22E

The Areaoftriangle is $11$

Explanation of Solution

Given:

The triangle points are $\left(-4,-5\right),\left(6,10\right),\left(6,-1\right)$

Concept used:

The area of triangle with vertices

  $\left(x_1,y_1\right),\left(x_2,y_2\right),\left(x_3,y_3\right)$

Area $=\pm \frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

Where the symbol $\pm$ indicates that the appropriate sign should be chosen to yield a positive area

Calculation:

The triangle points are given

  $\left(-4,-5\right),\left(6,10\right),\left(6,-1\right)$

  $\begin{array}{l}\left(x_1,y_1\right)=\left(-4,-5\right)\\ \left(x_2,y_2\right)=\left(6,10\right)\\ \left(x_3,y_3\right)=\left(6,-1\right)\end{array}$

The Areaof triangle

  $\begin{array}{l}\pm \frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|\\ \pm \frac{1}{2}\left|\begin{array}{ccc}-4& -5& 1\\ 6& 10& 1\\ 6& -1& 1\end{array}\right|\\ \pm \frac{1}{2}\left\{4\left(10+1\right)+5\left(6-6\right)+1\left(-6-60\right)\right\}\\ \pm \frac{1}{2}\left\{44+0-66\right\}\\ \pm \frac{1}{2}\left\{-22\right\}\\ 11\end{array}$