Chapter 13.3, Problem 13WE

a.

To determine

To show that EFGH is a rectangle.

Explanation of Solution

Given information: The points are,

  $R\left(-4,5\right)S\left(-1,9\right)T\left(7,3\right)U\left(4,-1\right)$ .

Proof: The required proof is obtained as,

  

  $\begin{array}{l}RS=\sqrt{{\left(-4-\left(-11\right)\right)}^2+{\left(9-5\right)}^2}=\sqrt{25}=5\\ ST=\sqrt{{\left(7-\left(-11\right)\right)}^2+{\left(3-9\right)}^2}=\sqrt{100}=10\\ TU=\sqrt{{\left(7-4\right)}^2+{\left(3+1\right)}^2}=\sqrt{25}=5\\ UR=\sqrt{{\left(4-\left(-4\right)\right)}^2+{\left(5+1\right)}^2}=\sqrt{100}=10\end{array}$

Therefore, $UR=ST$ and $RS=TU$ .

Now, we will check angle,

Slope of $RS=m_1=\frac{9-5}{-1-\left(-4\right)}=\frac{4}{3}$

Slope of $ST=m_2=\frac{3-9}{7-\left(-1\right)}=\frac{-6}{8}=-\frac{3}{4}$

  $m_1\times m_2=\frac{4}{3}\times -\frac{3}{4}=-1$

So, $\overline{RS}\perp \overline{ST}$ .

Hence this figure is rectangle.

b.

To determine

To verify that the diagonals are congruent by using the distance formula.

Explanation of Solution

Given information: The points are,

  $R\left(-4,5\right)S\left(-1,9\right)T\left(7,3\right)U\left(4,-1\right)$ .

Formula used: Distance formula.

The distance d between points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by:

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

Proof: The required solution is obtained as,

  

Diagonal $RT=\sqrt{{\left(7-\left(-4\right)\right)}^2+{\left(5-3\right)}^2}=\sqrt{121+4}=\sqrt{125}$

Diagonal $SU=\sqrt{{\left(4-\left(-1\right)\right)}^2+{\left(9-\left(-1\right)\right)}^2}=\sqrt{25+100}=\sqrt{125}$

So, $\overline{RT}$ and $\overline{SU}$ are equal in length

Hence, the diagonal are congruent.