Chapter 5.7, Problem 4PSA

To determine

Tograph:A quadrilateral formed byQ$\left(-11,5\right)$ , U$\left(7,5\right)$ ,A$\left(7,-13\right)$ and D$\left(-11,-13\right)$ .

Explanation of Solution

Given information:

The coordinates of point Q are $\left(-11,5\right)$ ,

The coordinates of point U are $\left(7,5\right)$ , The coordinates of point A are $\left(7,-13\right)$ , The coordinates of point Dare$\left(-11,-13\right)$ .

Graph:

  

Interpretation:

A quadrilateral formed bypointsQ$\left(-11,5\right)$ , U$\left(7,5\right)$ ,A$\left(7,-13\right)$ and D$\left(-11,-13\right)$ is square. All four angles are right angles and all sides are congruent. The sides can be calculated by applying distance formula.

The distance between two points by using the distance formula, which is an application of the Pythagoras theorem.

D$=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

  $(x_1,y_1)$ are coordinates of first point in the line.

  $(x_2,y_2)$ are coordinates of second point in the line.

  $\begin{array}{l}QU=\sqrt{{\left(7-\left(-11\right)\right)}^2+{\left(5-5\right)}^2}\\ QU=\sqrt{{\left(18\right)}^2+{\left(0\right)}^2}\\ QU=\sqrt{324}\\ QU=18\end{array}$

  $\begin{array}{l}QD=\sqrt{{\left(-11-\left(-11\right)\right)}^2+{\left(-13-5\right)}^2}\\ QD=\sqrt{{\left(0\right)}^2+{\left(-18\right)}^2}\\ QD=\sqrt{324}\\ QD=18\end{array}$

  $\begin{array}{l}UA=\sqrt{{\left(7-7\right)}^2+{\left(-13-5\right)}^2}\\ UA=\sqrt{{\left(0\right)}^2+{\left(-18\right)}^2}\\ UA=\sqrt{324}\\ UA=18\end{array}$

  $\begin{array}{l}AD=\sqrt{{\left(7-\left(-11\right)\right)}^2+{\left(-13-\left(-13\right)\right)}^2}\\ AD=\sqrt{{\left(18\right)}^2+{\left(0\right)}^2}\\ AD=\sqrt{324}\\ AD=18\end{array}$

QUAD is square because all sides are equal.