Chapter 13.3, Problem 12WE

a.

To determine

To show that EFGH is a rhombus.

Explanation of Solution

Given information: The points $E\left(-4,1\right),F\left(2,3\right),G\left(4,9\right),\text{ and }H\left(-2,7\right)$ .

Proof: The required proof is obtained as,

  

Distance of $EF=\sqrt{{\left(-4-2\right)}^2+{\left(1-3\right)}^2}=\sqrt{36+4}=\sqrt{40}$

Distance of $FG=\sqrt{{\left(2-4\right)}^2+{\left(3-9\right)}^2}=\sqrt{4+36}=\sqrt{40}$

Distance of $GH=\sqrt{{\left(4-\left(-2\right)\right)}^2+\left(9-7\right)}=\sqrt{36+4}=\sqrt{40}$

Distance of $HE=\sqrt{{\left(-4-\left(-2\right)\right)}^2+{\left(1-7\right)}^2}=\sqrt{4+36}=\sqrt{40}$

All sides are congruent.

Now, find the slopes,

  $\begin{array}{l}{m}_{EF}=\frac{3-1}{2-\left(-4\right)}=\frac{2}{6}=\frac{1}{3}\\ m_{FG}=\frac{9-3}{4-2}=\frac{6}{2}=3\\ m_{GH}=\frac{7-9}{-2-4}=\frac{-2}{-6}=\frac{1}{3}\\ m_{HE}=\frac{1-7}{-4-\left(-2\right)}=\frac{-6}{-2}=3\end{array}$

Both pairs of opposite sides are parallel.

Therefore, it is Rhombus.

Hence, EFGH is a rhombus.

b.

To determine

To show that diagonals are perpendicular, by using slopes.

Explanation of Solution

Given information: The points $E\left(-4,1\right),F\left(2,3\right),G\left(4,9\right),\text{ and }H\left(-2,7\right)$ .

Proof: The required proof is obtained as,

  

Slopes of diagonals,

  $\begin{array}{l}{m}_{EG}=\frac{9-1}{4-\left(-4\right)}=\frac{8}{8}=1\\ m_{FH}=\frac{7-3}{-2-2}=\frac{4}{-4}=-1\\ m_{EG}\times m_{FH}=-1\end{array}$

So, EG and FH are perpendicular.

Hence, the diagonals EG and FH are perpendicular.