Chapter 8.4, Problem 31E

(a)

To determine

To find: The sides that is perpendicular.

Answer to Problem 31E

  $y_1\&y_3$ are perpendicular to each other.

Explanation of Solution

Given information: $y_1=2x-6,y_2=-3x+4\&y_3=-\frac{x}{2}+4$

Calculation: The products of the slopes of perpendicular line is $-1$ .

Therefore, $y_1\&y_3$ are perpendicular.

(b)

To determine

To find: The vertices of the triangle.

Answer to Problem 31E

The vertices of the triangle are $\left(0,4\right),\left(4,2\right)\&\left(2,-2\right)$ .

Explanation of Solution

Given information: $y_1=2x-6,y_2=-3x+4\&y_3=-\frac{x}{2}+4$

Calculation:

  

From the graph, the vertices of the triangle are $\left(0,4\right),\left(4,2\right)\&\left(2,-2\right)$ .

(c)

To determine

To find: The perimeter and the area of the triangle.

Answer to Problem 31E

The perimeter and the area of the triangle are $15.27\&10$ , respectively.

Explanation of Solution

Given information: $y_1=2x-6,y_2=-3x+4\&y_3=-\frac{x}{2}+4$

Formula used: $A=\frac{b\times h}{2}$

Calculation: The perimeter of the triangle is

  $\begin{array}{l}=\sqrt{2^2+4^2}+\sqrt{4^2+2^2}+\sqrt{6^2+2^2}\\ =\sqrt{20}+\sqrt{20}+\sqrt{40}\\ =2\sqrt{20}+\sqrt{40}\\ =15.27\end{array}$

Now, the perimeter of the triangle:

  $\begin{array}{l}A=\frac{b\times h}{2}\\ A=\frac{1}{2}\times \sqrt{2^2+4^2}\times \sqrt{4^2+2^2}\\ A=\frac{1}{2}\times 20\\ A=10\end{array}$