Chapter 12.8, Problem 14E

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 14E

The value of $m_1=0$ , $m_2=\text{undefined}$ that is $\overline{DE}\perp \overline{EF}$ sot the triangle is the right trianglelength of each side is determined with the mid points.

Explanation of Solution

Given:

The given coordinate is $D\left(0,n\right),E\left(m,n\right),F\left(m,0\right)$ .

Calculation:

Consider the required triangle is shown in Figure 1

  

Figure 1

Consider the length $DE$ is,

  $\begin{array}{c}DE=\sqrt{{\left(m-0\right)}^2+{\left(n-n\right)}^2}\end{array}$

    $\begin{array}{c}=m\end{array}$

Consider the length $EF$ is,

  $\begin{array}{c}EF=\sqrt{{\left(m-m\right)}^2+{\left(0-n\right)}^2}\end{array}$

    $\begin{array}{c}=n\end{array}$

Consider the length $DF$ is,

  $\begin{array}{c}DF=\sqrt{{\left(m-0\right)}^2+{\left(0-n\right)}^2}\end{array}$

    $\begin{array}{c}=\sqrt{m^2+n^2}\end{array}$

Consider the slope of $DE$ is,

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

    $\begin{array}{c}=0\end{array}$

Consider the slope of $EF$ is,

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

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

Consider the slope of $DF$ is,

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

    $\begin{array}{c}=\frac{-n}{m}\end{array}$

Consider the midpoint of the side $DE$ is,

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

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

Consider the midpoint of the side $EF$ is,

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

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

Consider the midpoint of the side $DF$ is,

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

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

Since, the value of $m_1=0$ , $m_2=\text{undefined}$ that is $\overline{DE}\perp \overline{EF}$ sot the triangle is the right triangle.