Chapter 6.4, Problem 23PPS

To determine

The graph of each quadrilateral with given vertices.

Answer to Problem 23PPS

It is not a rectangle.

Explanation of Solution

Given:

The vertices are:

  $J\left(3,3\right),K\left(-5,2\right),L\left(-4,-4\right),M\left(4,-3\right)$

Calculation:

The graph will be:

The diagonals of a rectangle are congruent and bisect each other.

The length of JK is:

  $\begin{array}{l}JK=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ JK=\sqrt{{\left(-5-3\right)}^2+{\left(2-3\right)}^2}\\ JK=\sqrt{64+1}\\ JK=\sqrt{65}\end{array}$

The length of KL is:

  $\begin{array}{l}KL=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ KL=\sqrt{{\left(-4-\left(-5\right)\right)}^2+{\left(-4-2\right)}^2}\\ KL=\sqrt{1+36}\\ KL=\sqrt{37}\end{array}$

The length of LM is:

  $\begin{array}{l}LM=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ LM=\sqrt{{\left(4-\left(-4\right)\right)}^2+{\left(-3-\left(-4\right)\right)}^2}\\ LM=\sqrt{64+1}\\ LM=\sqrt{65}\end{array}$

The length of JM is:

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

As, JK = LM and KL = JM, the opposite sides are congruent.

Hence, QRST is a parallelogram.

Also, when the diagonals of a parallelogram are congruent, then it is a rectangle.

The length of KM is:

  $\begin{array}{l}KM=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ KM=\sqrt{{\left(4-\left(-5\right)\right)}^2+{\left(-3-2\right)}^2}\\ KM=\sqrt{81+25}\\ KM=\sqrt{106}\end{array}$

The length of JL is:

  $\begin{array}{l}JL=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ JL=\sqrt{{\left(-4-3\right)}^2+{\left(-4-3\right)}^2}\\ JL=\sqrt{49+49}\\ JL=\sqrt{98}\end{array}$

As, KM is not equal to JL, the diagonals are not congruent.

Hence, it is not a rectangle.