Chapter 13.6, Problem 35WE

a.

To determine

To graph: on the same axes.

Explanation of Solution

Given information: The equations are $y=\frac{1}{2}x-2,y=-2x+3\&y=3x+8$ .

Graph:

The graph will be:

  

Interpretation:

From the above graph it is observed that the blue line represents $y=3x+8$ , the green line represent $y=-2x+3$ and the black line represent $y=\frac{1}{2}x-2$ .

b.

To determine

To find: the coordinate where the lines intersect.

Answer to Problem 35WE

The coordinates are $\left(2,-1\right)\left(-1,5\right)\&\left(-4,-4\right)$ .

Explanation of Solution

Given information: The equations are $y=\frac{1}{2}x-2,y=-2x+3\&y=3x+8$ .

Calculation:

The intersection of $y=\frac{1}{2}x-2\&y=-2x+3$ is $\left(2,-1\right)$ .

The intersection of $y=\frac{1}{2}x-2\&y=3x+8$ is $\left(-1,5\right)$ .

The intersection of $y=-2x+3\&y=3x+8$ is $\left(-4,-4\right)$ .

c.

To determine

To find: the area of the triangle.

Answer to Problem 35WE

The area of triangle is $\frac{45}{2}$ .

Explanation of Solution

Given information: The equations are $y=\frac{1}{2}x-2,y=-2x+3\&y=3x+8$ .

Calculation:

  $y=\frac{1}{2}x-2\&y=-2x+3$ are perpendicular.

So, it’s a right angle triangle.

Base $=\sqrt{{\left(2+1\right)}^2+{\left(5+1\right)}^2}=\sqrt{3^2+6^2}=\sqrt{45}$

Height $=\sqrt{{\left(2+4\right)}^2+{\left(4-1\right)}^2}=\sqrt{6^2+3^2}=\sqrt{45}$

Area $=\frac{1}{2}\times \text{base}\times \text{height}$

  $\begin{array}{l}\Rightarrow =\frac{1}{2}\times \sqrt{45}\times \sqrt{45}\\ \Rightarrow \frac{45}{2}\end{array}$