Chapter 13.4, Problem 11MRE

a.

To determine

To prove: That $\Delta DEF$ is a right triangle.

Explanation of Solution

Given information: Triangle DEF has vertices $D\left(-5,1\right),E\left(-2,-3\right),F\left(6,3\right)$ .

Formula used: Distance between any two points having points $\left(x_1,y_1\right)$ two $\left(x_2,y_2\right)$ is

  $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$ .

Proof:

Distance between any two points having points $\left(x_1,y_1\right)$ two $\left(x_2,y_2\right)$ is

  $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$ .

Triangle DEF has vertices $D\left(-5,1\right),E\left(-2,-3\right),F\left(6,3\right)$ .

Then,

  $\begin{array}{l}DE=\sqrt{{\left(-5+2\right)}^2+{\left(1+3\right)}^2}\\ DE=\sqrt{25}\\ DE=5\end{array}$

  $\begin{array}{l}EF=\sqrt{{\left(-2-6\right)}^2+{\left(-3-3\right)}^2}\\ EF=\sqrt{100}\\ EF=10\\ DF=\sqrt{{\left(-5-6\right)}^2+{\left(1-3\right)}^2}\\ DF=\sqrt{125}\\ DF=5\sqrt{5}\end{array}$

The Pythagorean theorem tells us that the relationship in every triangle is

  $D{F}^2=E{F}^2+D{E}^2$

So, $\Delta DEF$ is a right triangle.

b.

To determine

To prove: That $\Delta DEF$ is a right triangle.

Explanation of Solution

Given information: Triangle DEF has vertices $D\left(-5,1\right),E\left(-2,-3\right),F\left(6,3\right)$ .

Formula used: For the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ , slope will be

  $\text{slope = }\frac{y_2-y_2}{x_2-x_1}$ .

Proof:

For the points $\left(x_1,y_1\right)$ two $\left(x_2,y_2\right)$ , slope will be

  $\text{slope = }\frac{y_2-y_2}{x_2-x_1}$ .

Triangle DEF has vertices $D\left(-5,1\right),E\left(-2,-3\right),F\left(6,3\right)$ .

For the points $D\left(-5,1\right)$ and $E\left(-2,-3\right)$ , slope will be

  $\begin{array}{l}{m}_1=\frac{-3-1}{-2+5}\\ m_1=\frac{-4}{3}\end{array}$

For the points $F\left(6,3\right)$ and $E\left(-2,-3\right)$ , slope will be

  $\begin{array}{l}{m}_2=\frac{-3-3}{-2-6}\\ m_2=\frac{3}{4}\end{array}$

And $m_1\cdot m_2=-1$

So, $\Delta DEF$ is a right triangle.