Chapter 11.3, Problem 59AYU

To determine

To find: The area of a triangle with vertices $(2,3),(5,2)\&(6,5).$

Answer to Problem 59AYU

  $\text{5}$.

Explanation of Solution

Given information:

A triangle has vertices $(x_1,y_1),(x_2,y_2)\&(x_3,y_3).$

The area of the triangle is given by the absolute value of $\text{D,}$ where,

  $\text{D=}\frac{1}{2}\left|\begin{array}{l}{\text{x}}_1{\text{ x}}_2{\text{ x}}_3\\ {\text{y}}_1{\text{y}}_2{\text{ y}}_3\\ 11\text{ 1}\end{array}\right|$

Since, The vertices of the triangle $(2,3),(5,2)\&(6,5)$ .

Therefore,

  $\begin{array}{l}{\text{x}}_1=2,{\text{y}}_2=3\\ {\text{x}}_2=5,{\text{y}}_2=2\\ {\text{x}}_3=6,{\text{y}}_3=5\end{array}$

Let’s find the area of the triangle

  $\begin{array}{l}\text{D=}\frac{1}{2}\left|\begin{array}{l}2\text{ 5 6}\\ \text{3 2 5}\\ \text{1 1 1 }\end{array}\right|\\ \text{ =}\frac{1}{2}\left\{2(2-5)-5(3-5)+6(3-2)\right\}\\ \text{ =}\frac{1}{2}\left\{2(-3)-5(-2)+6\right\}\\ \text{ =}\frac{1}{2}(-6+10+6)\\ \text{ =5}\end{array}$

Hence, The area of the triangle is $5$