Chapter 13.3, Problem 14WE

a.

To determine

To show that NOPG is an isosceles trapezoid.

Explanation of Solution

Given information: The points are,

  $N\left(-1,-5\right)\text{ O}\left(0,0\right)\text{ P}\left(3,2\right)\text{ Q}\left(8,1\right)$ .

Proof: The required proof is obtained as,

  

  $\begin{array}{l}NO=\sqrt{{\left(-1\right)}^2+{\left(-5\right)}^2}=\sqrt{26}\\ PQ=\sqrt{{\left(5\right)}^2+{\left(1\right)}^2}=\sqrt{26}\\ PO=\sqrt{{\left(3\right)}^2+{\left(2\right)}^2}=\sqrt{145}\\ QN=\sqrt{{\left(9\right)}^2+{\left(6\right)}^2}=\sqrt{145}\end{array}$

Now, we will check angle,

Slope of $PO=\frac{2-0}{3-0}=\frac{2}{3}$

Slope of $QN=\frac{1-\left(-5\right)}{8-\left(-1\right)}=\frac{6}{9}=\frac{2}{3}$

Therefore, $m_{PO}=m_{QN}$ and $NO=PQ$ .

Hence this is an Isosceles Trapezoid.

b.

To determine

To show that the diagonals are congruent.

Explanation of Solution

Given information: The points are,

  $N\left(-1,-5\right)\text{ O}\left(0,0\right)\text{ P}\left(3,2\right)\text{ Q}\left(8,1\right)$ .

Proof: The required solution is obtained as,

Diagonal $PN=\sqrt{{\left(3-\left(-1\right)\right)}^2+{\left(2+5\right)}^2}=\sqrt{16+49}=\sqrt{65}$

Diagonal $OQ=\sqrt{{\left(8-0\right)}^2+{\left(1-0\right)}^2}=\sqrt{64+1}=\sqrt{65}$

So, $PN$ and $OQ$ are equal in length

Hence, the diagonal are congruent.