Chapter 6.3, Problem 25PPS

To determine

The graph of each quadrilateral with given vertices and whether the figure forms a parallelogram or not.

Answer to Problem 25PPS

JKLM is not a parallelogram.

Explanation of Solution

Given:

The given vertices are:

  $J\left(-4,-4\right),K\left(-3,1\right),L\left(4,3\right),M\left(3,-3\right).$

Calculation:

A quadrilateral is a parallelogram if both pairs of its opposite sides are congruent.

The quadrilateral with given vertices will be:

The length of JK is:

  $\begin{array}{l}JK=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ JK=\sqrt{{\left(-3-\left(-4\right)\right)}^2+{\left(1-\left(-4\right)\right)}^2}\\ JK=\sqrt{1+25}\\ JK=\sqrt{26}\end{array}$

The length of KL is:

  $\begin{array}{l}KL=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ KL=\sqrt{{\left(4-\left(-3\right)\right)}^2+{\left(3-1\right)}^2}\\ KL=\sqrt{49+4}\\ KL=\sqrt{53}\end{array}$

The length of LM is:

  $\begin{array}{l}LM=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ LM=\sqrt{{\left(3-4\right)}^2+{\left(-3-3\right)}^2}\\ LM=\sqrt{1+36}\\ LM=\sqrt{37}\end{array}$

The length of JM is:

  $\begin{array}{l}JM=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ JM=\sqrt{{\left(3-\left(-4\right)\right)}^2+{\left(-3-\left(-4\right)\right)}^2}\\ JM=\sqrt{49+1}\\ JM=\sqrt{50}\end{array}$

As JK is not equal to ML and JM is not equal to KL, therefore none of the opposite sides are congruent.

Therefore, by the definition of a parallelogram, JKLM is not a parallelogram.