Chapter 4.3, Problem 17PPS

To determine

To find: the whether the given quadrilateral $ABCD$ is a trapezoidal or not.

Answer to Problem 17PPS

The given quadrilateral is a trapezoidal.

Slope of $BC=AD\text{and}BA\ne CD$

Explanation of Solution

Given information:

A trapezoid is a quadrilateral that has exactly one pair of parallel opposite sides.

Given graph,

  

Calculation:

  $BC$ and $AD$ line segment slope need to be calculated.

Coordinate point $B$ is $\left(0,3\right)$

Coordinate point $C$ is $\left(3,4\right)$

Coordinate point $A$ is $\left(0,-3\right)$

Coordinate point $D$ is $\left(6,-1\right)$

The slope of the line

  $\begin{array}{c}BC\left(m_1\right)=\frac{y_2-y_1}{x_2-x_1}\\ m_1=\frac{4-3}{3-0}\\ =\frac{1}{3}\end{array}$

The slope of the line

  $\begin{array}{c}AD\left(m_2\right)=\frac{y_2-y_1}{x_2-x_1}\\ m_2=\frac{-1-\left(-3\right)}{6-0}\\ =\frac{1}{3}\end{array}$

Slope of $BC=AD$

  $BA$ And $CD$ line segment slope need to be calculated.

The slope of the line

  $\begin{array}{c}BA\left(m_3\right)=\frac{y_2-y_1}{x_2-x_1}\\ m_3=\frac{-3-3}{0-0}\\ =\infty \end{array}$

The slope of the line

  $\begin{array}{c}CD\left(m_4\right)=\frac{y_2-y_1}{x_2-x_1}\\ m_4=\frac{-1-4}{6-3}\\ =-\frac{5}{3}\end{array}$

Since exactly a pair of lines are parallel to each other,

The given quadrilateral is a trapezoidal.

Slope of $BC=AD\text{and}BA\ne CD$