Chapter 1.5, Problem 11CFU

To determine

whether $EFGH$ is a parallelogram, a rectangle, or neither.

Answer to Problem 11CFU

Parallelogram

Explanation of Solution

Given information:

If both pairs of its opposite side are parallel, then the quadrilateral is a parallelogram.

A parallelogram is a rectangle if its adjacent sides are perpendicular.

The given quadrilateral is shown below.

  

From the graph, $E\left(3,4\right),F\left(4,8\right),G\left(7,6\right),H\left(6,2\right)$ .

  $\begin{array}{l}\overline{EF}\text{ is opposite to }\overline{GH}\\ \overline{EH}\text{ is opposite to }\overline{GH}\end{array}$

Calculation:

The formula of slope is as follows:

  $m=\frac{y_2-y_1}{x_2-x_1}$

Substitute the required values in the slope formula to evaluate the slope for $\overline{EF}$ as shown below.

  $\begin{array}{c}m=\frac{8-4}{4-3}\\ =\frac{4}{1}\\ =4\end{array}$

Substitute the required values in the slope formula to evaluate the slope for $\overline{GH}$ as shown below.

  $\begin{array}{c}m=\frac{2-6}{6-7}\\ =\frac{-4}{-1}\\ =4\end{array}$

Substitute the required values in the slope formula to evaluate the slope for $\overline{EF}$ as shown below.

  $\begin{array}{c}m=\frac{8-4}{4-3}\\ =\frac{4}{1}\\ =4\end{array}$

Since, the slope of $\overline{EF}$ and $\overline{GH}$ are equal. So, $\overline{EF}\parallel \overline{GH}$ .

Substitute the required values in the slope formula to evaluate the slope for $\overline{EH}$ as shown below.

  $\begin{array}{c}m=\frac{2-4}{6-3}\\ =\frac{-2}{3}\end{array}$

Substitute the required values in the slope formula to evaluate the slope for $\overline{FG}$ as shown below.

  $\begin{array}{c}m=\frac{6-8}{7-4}\\ =\frac{-2}{3}\end{array}$

Since, the slope of $\overline{EH}$ and $\overline{FG}$ are equal. So, $\overline{EH}\parallel \overline{FG}$ .

Since, both pairs of the opposite sides are parallel. Thus, $EFGH$ is a parallelogram.

Since, the slope of $\overline{EF}$ and $\overline{FG}$ is not perpendicular because the slopes are opposite reciprocal of each other.

  $\begin{array}{l}\perp m=-\frac{1}{m}\\ 4=-\frac{1}{\left(-\frac{2}{3}\right)}\\ 4\ne \frac{3}{2}\end{array}$

Thus, the $EFGH$ is not a rectangle.