Chapter 6.6, Problem 78SPR

To determine

The graph of each quadrilateral with given vertices and whether it is a rectangle using Distance Formula.

Answer to Problem 78SPR

  

ABCD is not a rectangle.

Explanation of Solution

Given:

The given vertices are:

  $A\left(4,2\right),B\left(-4,1\right),C(-3,-5),D\left(5,-4\right).$

Calculation:

Draw the quadrilateral using given vertices:

  

The length of AB:

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

The length of BC:

  $\begin{array}{l}BC=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ BC=\sqrt{{\left(-3-\left(-4\right)\right)}^2+{\left(-5-1\right)}^2}\\ BC=\sqrt{1+36}\\ BC=\sqrt{37}\end{array}$

The length of CD:

  $\begin{array}{l}CD=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ CD=\sqrt{{\left(5-\left(-3\right)\right)}^2+{\left(-4-\left(-5\right)\right)}^2}\\ CD=\sqrt{64+1}\\ CD=\sqrt{65}\end{array}$

The length of DA:

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

Hence, the opposite sides are congruent.

Therefore, ABCD is a parallelogram.

For a parallelogram to be a rectangle, its diagonals are congruent.

The length of AC will be:

  $\begin{array}{l}AC=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ AC=\sqrt{{\left(-3-4\right)}^2+{\left(-5-2\right)}^2}\\ AC=\sqrt{49+49}\\ AC=\sqrt{98}\end{array}$

The length of BD will be:

  $\begin{array}{l}BD=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ BD=\sqrt{{\left(5-\left(-4\right)\right)}^2+{\left(-4-1\right)}^2}\\ BD=\sqrt{25+81}\\ BD=\sqrt{106}\end{array}$

As AC is not equal to BD, the diagonals are not congruent.

Therefore, ABCD is not a rectangle.