Chapter 12.8, Problem 13E

To determine

To find: The graph of the triangle with the given vertices and then length and the slope of the side of the triangle and then determine the coordinates of the midpoints of each side. Then find if the triangle is right triangle or right triangle or isosceles.

Answer to Problem 13E

The triangle is the triangle isosceles triangle the graph is shown in Figure 1, the length of each side is determined with the mid points.

Explanation of Solution

Given:

The given coordinate is $A\left(0,0\right),B\left(h,h\right),C\left(2h,0\right)$ .

Calculation:

Consider the required triangle is shown in Figure 1

  

Figure 1

Consider the length $AB$ is,

  $\begin{array}{c}AB=\sqrt{h^2+h^2}\end{array}$

    $\begin{array}{c}=\sqrt{2h}\end{array}$

Consider the length $BC$ is,

  $\begin{array}{c}BC=\sqrt{{\left(2h-0\right)}^2+{\left(0-h\right)}^2}\end{array}$

    $\begin{array}{c}=\sqrt{2h}\end{array}$

Consider the length $AC$ is,

  $\begin{array}{c}AC=\sqrt{{\left(2h-0\right)}^2+{\left(0-0\right)}^2}\end{array}$

    $\begin{array}{c}=2h\end{array}$

Consider the slope of $AB$ is,

  $\begin{array}{c}{m}_1=\frac{h-0}{h-0}\end{array}$

    $\begin{array}{c}=1\end{array}$

Consider the slope of $BC$ is,

  $\begin{array}{c}{m}_2=\frac{0-h}{2h-h}\end{array}$

    $\begin{array}{c}=-1\end{array}$

Consider the slope of $AC$ is,

  $\begin{array}{c}{m}_3=\frac{2h-0}{0-0}\end{array}$

    $\begin{array}{c}=\text{undefined}\end{array}$

Consider the midpoint of the side $AB$ is,

  $\begin{array}{c}{M}_1=\left(\frac{0+h}{2},\frac{0+h}{2}\right)\end{array}$

    $\begin{array}{c}=\left(\frac{h}{2},\frac{h}{2}\right)\end{array}$

Consider the midpoint of the side $BC$ is,

  $\begin{array}{c}{M}_2=\left(\frac{2h+h}{2},\frac{h+0}{2}\right)\end{array}$

    $\begin{array}{c}=\left(\frac{3h}{2},\frac{h}{2}\right)\end{array}$

Consider the midpoint of the side $AC$ is,

  $\begin{array}{c}{M}_3=\left(\frac{0+2h}{2},\frac{0+0}{2}\right)\end{array}$

    $\begin{array}{c}=\left(h,0\right)\end{array}$

Since, the line AB is equal to BC, $m_1\cdot m_2=-1$ , then $AB\perp BC$ so the triangle is the triangle isosceles triangle.