Chapter 11.3, Problem 59AYU

Geometry: Area of a Triangle A triangle has vertices $\left(x_1,x_2\right),\left(x_2,y_2\right),\text{and}\left(x_3,y_3\right)$. The area of the triangle is given by the absolute value of $D$, where $D=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& x_2& x_3\\ y_1& y_2& y_3\\ 1& 1& 1\end{array}\right|$.

Use this formula to find the area o9f a triangle with vertices $\left(2,3_2\right),\left(5,2\right),\text{and}\left(6,5\right)$.

To determine

To find: A triangle has vertices $\left(x_1,y_1\right)$, $\left(x_2,y_2\right)$ and $\left(x_3,y_3\right)$. The area of the triangle is given by the absolute value of $D$ where

$D = \frac{1}{2}\left|\begin{array}{lll}{x}_1\hfill & x_2\hfill & x_3\hfill \\ y_1\hfill & y_2\hfill & y_3\hfill \\ 1\hfill & 1\hfill & 1\hfill \end{array}\right|$

Use this formula to find the area of a triangle with vertices $\left(2,3\right)$, $\left(5,2\right)$ and $\left(6,5\right)$.

Answer to Problem 59AYU

$D = 5$ Square units

The area of a triangle with vertices $\left(2,3\right)$, $\left(5,2\right)$ and $\left(6,5\right)$ is 5 square units.

Explanation of Solution

Given:

Triangle with vertices $\left(2,3\right)$, $\left(5,2\right)$ and $\left(6,5\right)$.

Calculation:

$\left(2,3\right)$, $\left(5,2\right)$ and $\left(6,5\right)$.

$D = \frac{1}{2}\left|\begin{array}{lll}{x}_1\hfill & x_2\hfill & x_3\hfill \\ y_1\hfill & y_2\hfill & y_3\hfill \\ 1\hfill & 1\hfill & 1\hfill \end{array}\right|$

We have

$D = \frac{1}{2}\left|\begin{array}{lll}2\hfill & 5\hfill & 6\hfill \\ 3\hfill & 2\hfill & 5\hfill \\ 1\hfill & 1\hfill & 1\hfill \end{array}\right|$

$D=\frac{1}{2}\left[2\left|\begin{array}{c}2\\ 1\end{array} \begin{array}{c}5\\ 1\end{array}\right|-5\left|\begin{array}{c}3\\ 1\end{array} \begin{array}{c}5\\ 1\end{array}\right|+6\left|\begin{array}{c}3\\ 1\end{array} \begin{array}{c}2\\ 1\end{array}\right|\right]$

$D=\frac{1}{2}\left[2\left(2-5\right)-5\left(3-5\right)+6\left(3-2\right)\right]$

$D = 5$ Square units