Chapter 6.6, Problem 14PPS

To determine

Whether the figure with given vertices is a trapezoid.

Answer to Problem 14PPS

QRST is a trapezoid.

The lengths are equal, hence QRST is an isosceles trapezoid.

Explanation of Solution

Given:

The given vertices are:

  $Q\left(2,5\right),R\left(-2,1\right),S(-1,-6),T\left(9,4\right).$

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 QR will be:

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

The slope of RS will be:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{-6-1}{-1+2}\\ m=\frac{-7}{1}\\ m=-7\end{array}$

The slope of ST will be:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{4-\left(-6\right)}{9-\left(-1\right)}\\ m=\frac{10}{10}\\ m=1\end{array}$

The slope of TQ will be:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{5-4}{2-9}\\ m=\frac{1}{-7}\\ m=-\frac{1}{7}\end{array}$

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

Hence, QRST is a trapezoid.

Now, apply the Distance Formula to compare the lengths:

The length of RS will be:

  $\begin{array}{l}RS=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ RS=\sqrt{{\left(-1-\left(-2\right)\right)}^2+{\left(-6-1\right)}^2}\\ RS=\sqrt{1+49}\\ RS=\sqrt{50}\end{array}$

The length of TQ will be:

  $\begin{array}{l}TQ=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ TQ=\sqrt{{\left(2-9\right)}^2+{\left(5-4\right)}^2}\\ TQ=\sqrt{49+1}\\ TQ=\sqrt{50}\end{array}$

The lengths are equal, hence QRST is an isosceles trapezoid.