Chapter 12.3, Problem 5PSA

a.

To determine

To find: The slopes of the diagonals $\overline{AC}\text{and}\overline{BD}$ of a parallelogram $ABCD.$

Answer to Problem 5PSA

The slopes of the diagonals $\overline{AC}\text{and}\overline{BD}$ are $-2\text{and}\frac{1}{2}$ respectively.

Explanation of Solution

Given Information:

  $A=\left(3,6\right),B=\left(13,6\right),C=\left(7,-2\right)\text{and}D=\left(-3,-2\right)$

Formula used:

Slope of a line segment with endpoints $\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)$ is

  $m=\frac{y_2-y_1}{x_2-x_1}$

Calculation:

The coordinates of four points of parallelogram are given as

  $A=\left(3,6\right),B=\left(13,6\right),C=\left(7,-2\right)\text{and}D=\left(-3,-2\right)$

We know that, slope of a line with end points $\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)$ is given by

  $m=\frac{y_2-y_1}{x_2-x_1}$

Let $m_1\text{and}{m}_2$ be the slopes of $\overline{AC}\text{and}\overline{BD}$ respectively

  $\begin{array}{l}\therefore m_1=\frac{-2-6}{7-3}\\ m_1=\frac{-8}{4}\\ m_1=-2\end{array}$

And, $m_2=\frac{-2-6}{-3-13}$

  $\begin{array}{l}{m}_2=\frac{-8}{-16}\\ m_2=\frac{1}{2}\end{array}$

Hence, the slopes of the diagonals $\overline{AC}\text{and}\overline{BD}$ are $-2\text{and}\frac{1}{2}$ respectively.

b.

To determine

To find: the specific name of the parallelogram $ABCD.$

Answer to Problem 5PSA

The parallelogram $ABCD$ is a rhombus.

Explanation of Solution

Given Information:

From part a,

Slope of $\overline{AC}\text{is}\left(m_1\right)=-2$

Slope of $\overline{BD}\text{is}\left(m_2\right)=\frac{1}{2}$

It is given that

Slope of $\overline{AC}\text{is}\left(m_1\right)=-2$

Slope of $\overline{BD}\text{is}\left(m_2\right)=\frac{1}{2}$

Now, is we take product of the slopes then

  $\begin{array}{l}{m}_1\times m_2=-2\times \frac{1}{2}\\ \therefore m_1\times m_2=-1\end{array}$

Thus, the product of the slopes of two lines is $-1.$

We know that, when the product of slopes is $-1$ , then the lines are perpendicular.

So, the diagonals $\overline{AC}\text{and}\overline{BD}$ are perpendicular to each other.

Hence, the parallelogram $ABCD$ is a rhombus.