Chapter 6.4, Problem 24PPS

To determine

The graph of each quadrilateral with given vertices.

Answer to Problem 24PPS

It is a rectangle.

Explanation of Solution

Given:

The vertices are:

  $Q\left(-2,2\right),R\left(0,-2\right),S\left(6,1\right),T\left(4,5\right)$

Calculation:

The graph will be:

The opposite sides of a parallelogram are congruent.

The length of QR is:

  $\begin{array}{l}QR=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ QR=\sqrt{{\left(0-\left(-2\right)\right)}^2+{\left(-2-2\right)}^2}\\ QR=\sqrt{4+16}\\ QR=\sqrt{20}\end{array}$

The length of RS is:

  $\begin{array}{l}RS=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ RS=\sqrt{{\left(6-0\right)}^2+{\left(1-\left(-2\right)\right)}^2}\\ RS=\sqrt{36+9}\\ RS=\sqrt{45}\end{array}$

The length of ST is:

  $\begin{array}{l}ST=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ ST=\sqrt{{\left(4-6\right)}^2+{\left(5-1\right)}^2}\\ ST=\sqrt{4+16}\\ ST=\sqrt{20}\end{array}$

The length of QT is:

  $\begin{array}{l}QT=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ QT=\sqrt{{\left(4-\left(-2\right)\right)}^2+{\left(5-2\right)}^2}\\ QT=\sqrt{36+9}\\ QT=\sqrt{45}\end{array}$

As, QR = ST and RS = QT, 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 QS is:

  $\begin{array}{l}QS=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ QS=\sqrt{{\left(6-\left(-2\right)\right)}^2+{\left(2-1\right)}^2}\\ QS=\sqrt{64+1}\\ QS=\sqrt{65}\end{array}$

The length of RT is:

  $\begin{array}{l}RT=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ RT=\sqrt{{\left(4-0\right)}^2+{\left(5-\left(-2\right)\right)}^2}\\ RT=\sqrt{16+49}\\ RT=\sqrt{65}\end{array}$

As, QS = RT, the diagonals are congruent.

Hence, it is a rectangle.