Chapter 6.6, Problem 60PPS

To determine

To calculate:

The figure represented by the following coordinates is trapezoid, parallelogram, square, rhombus or quadrilateral:

  $\mathrm{A}\left(-1,4\right),\mathrm{B}\left(2,6\right),\mathrm{C}\left(3,3\right),\mathrm{D}\left(0,1\right)$

Answer to Problem 60PPS

The figure is parallelogram

Explanation of Solution

Given information:

  $\mathrm{A}\left(-1,4\right),\mathrm{B}\left(2,6\right),\mathrm{C}\left(3,3\right),\mathrm{D}\left(0,1\right)$

Calculation:

First graph all the points as:

  

Now, calculate the slopes of all sides:

  $\begin{array}{l}{\mathrm{m}}_{\mathrm{A}\mathrm{B}}=\frac{6-4}{2-\left(-1\right)}=\frac{2}{3}\\ {\mathrm{m}}_{\mathrm{B}\mathrm{C}}=\frac{3-6}{3-1}=-3\\ {\mathrm{m}}_{\mathrm{C}\mathrm{D}}=\frac{1-3}{0-3}=\frac{2}{3}\\ {\mathrm{m}}_{\mathrm{B}\mathrm{C}}=\frac{1-4}{0-\left(-1\right)}=-3\end{array}$

Here, slopes of each pair opposite sides are equal so, one can say that two pairs of the opposite sides are parallel which results that the quadrilateral is parallelogram. Angles are not right angles because none of adjacent sides have the slopes of whose product is $-1$ .

Now, calculate the length of the parallelograms by using the distance formula:

  $\begin{array}{l}\mathrm{A}\mathrm{B}=\sqrt{{\left(2-\left(-1\right)\right)}^2+{\left(6-4\right)}^2}=\sqrt{13}\\ \mathrm{B}\mathrm{C}=\sqrt{{\left(3-2\right)}^2+{\left(3-6\right)}^2}=\sqrt{10}\\ \mathrm{C}\mathrm{D}=\sqrt{{\left(0-3\right)}^2+{\left(1-3\right)}^2}=\sqrt{13}\\ \mathrm{A}\mathrm{D}=\sqrt{{\left(0-\left(-1\right)\right)}^2+{\left(1-4\right)}^2}=\sqrt{10}\end{array}$

The quadrilateral is parallelogram as no consecutive sides are congruent.