Chapter 13.3, Problem 17PSC

To determine

To find: The area of the triangle whose sides lie on the graph of given lines.

Answer to Problem 17PSC

The area is $11$ square unit.

Explanation of Solution

Given:

The given equations of given lines are:

  $\begin{array}{c}3x+y+1=0\\ x+4y-7=0\\ -5x+2y+13=0\end{array}$ .

Calculation:

The graph of the above equations is given below:

  

From the graph it is clear, $\left(3,1\right),\left(-1,2\right),\left(1,-4\right)$ are the vertices of the triangle.

The area of triangle with the three vertices is calculated as:

  $\begin{array}{c}\text{Area}=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|\\ =\frac{1}{2}\left|3\left(2-\left(-4\right)\right)+\left(-1\right)\left(-4-1\right)+1\left(1-2\right)\right|\\ =11\end{array}$

Therefore the area is $11$ square unit.