Chapter 6.6, Problem 15PPS

To determine

Whether the figure with given vertices is a trapezoid.

Answer to Problem 15PPS

WXYZ is a trapezoid.

The lengths are not equal, hence WXYZ is not an isosceles trapezoid.

Explanation of Solution

Given:

The given vertices are:

  $W\left(-5,-1\right),X\left(-2,2\right),Y(3,1),Z\left(5,-3\right).$

Calculation:

Draw the quadrilateral using given vertices:

  

A trapezoid has exactly one pair of opposite sides.

Use the slope formula to compare the opposite sides:

The slope of WX will be:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{2-\left(-1\right)}{-2-\left(-5\right)}\\ m=\frac{3}{3}\\ m=1\end{array}$

The slope of XY will be:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{1-2}{3-\left(-2\right)}\\ m=\frac{-1}{5}\\ m=-\frac{1}{5}\end{array}$

The slope of YZ will be:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{-3-1}{5-3}\\ m=\frac{-4}{2}\\ m=-2\end{array}$

The slope of ZW will be:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{-1-\left(-3\right)}{-5-5}\\ m=\frac{2}{-10}\\ m=-\frac{1}{5}\end{array}$

The slopes of exactly one pair of opposite sides are equal, so exactly one pair of opposite sides is parallel.

Hence, WXYZ is a trapezoid.

Now, apply the Distance Formula to compare the lengths:

The length of WX will be:

  $\begin{array}{l}WX=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ WX=\sqrt{{\left(-2-\left(-5\right)\right)}^2+{\left(2-\left(-1\right)\right)}^2}\\ WX=\sqrt{9+9}\\ WX=\sqrt{18}\end{array}$

The length of YZ will be:

  $\begin{array}{l}YZ=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ YZ=\sqrt{{\left(5-3\right)}^2+{\left(-3-1\right)}^2}\\ YZ=\sqrt{4+16}\\ YZ=\sqrt{20}\end{array}$

The lengths are not equal, hence WXYZ is not an isosceles trapezoid.