Chapter 6.6, Problem 12PPS

To determine

Whether the figure with given vertices is a trapezoid.

Answer to Problem 12PPS

ABCD is a trapezoid.

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

Explanation of Solution

Given:

The given vertices are:

  $A\left(-2,5\right),B\left(-3,1\right),C(6,1),D\left(3,5\right).$

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 AB will be:

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

The slope of BC will be:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{3-3}{3-\left(-2\right)}\\ m=\frac{0}{5}\\ m=0\end{array}$

The slope of CD will be:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{5-1}{3-6}\\ m=\frac{4}{-3}\\ m=-\frac{4}{3}\end{array}$

The slope of AD will be:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{-1-\left(-1\right)}{-4-5}\\ m=\frac{0}{-9}\\ m=0\end{array}$

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

Hence, ABCD is a trapezoid.

Now, apply the Distance Formula to compare the lengths:

The length of AB will be:

  $\begin{array}{l}AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ AB=\sqrt{{\left(-3-\left(-2\right)\right)}^2+{\left(1-5\right)}^2}\\ AB=\sqrt{1+16}\\ AB=\sqrt{17}\end{array}$

The length of CD will be:

  $\begin{array}{l}CD=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ CD=\sqrt{{\left(3-6\right)}^2+{\left(5-1\right)}^2}\\ CD=\sqrt{9+16}\\ CD=\sqrt{25}\end{array}$

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