Chapter 1.1, Problem 55E

To determine

To find:The coordinates of missing vertices of three different parallelograms and their diagram.

Answer to Problem 55E

The missing vertices of three different parallelograms with three common vertices are $\left(5,2\right)$ , $\left(-1,4\right)$ and $\left(-1,-2\right)$ .

Explanation of Solution

Given information:The vertices of three different parallelograms are $\left(-1,1\right)$ , $\left(2,0\right)$ and $\left(2,3\right)$ .

Calculation:

Let the three vertices of different parallelograms are $P\left(-1,1\right)$ , $Q\left(2,0\right)$ and $R\left(2,3\right)$ .

Case 1: Consider two sides of parallelogram as $\overline{PQ}$ and $\overline{QR}$ .

The line $\overline{QR}$ corresponds to the line $x=2$ . So, the side parallel to $\overline{QR}$ will be $x=-1$ .

To find the line parallel to $\overline{PQ}$ , find the slope of line $\overline{PQ}$ .

  $\begin{array}{c}{m}_{PQ}=\frac{0-1}{2-\left(-1\right)}\\ =\frac{-1}{3}\end{array}$

The equation of line parallel to $\overline{PQ}$ is:

  $\begin{array}{c}y=m\left(x-x_1\right)+y_1\\ y-3=-\frac{1}{3}\left(x-2\right)\\ y=-\frac{1}{3}\left(x-2\right)+3\end{array}$

To find the fourth vertex, substitute $-1$ for $x$ in the above equation.

  $\begin{array}{l}y=-\frac{1}{3}\left(-1-2\right)+3\\ y=1+3\\ y=4\end{array}$

So, the fourth vertex of the parallelogram is $\left(-1,4\right)$ . The parallelogram with vertices $\left(-1,1\right)$ , $\left(2,0\right)$ , $\left(2,3\right)$ and $\left(-1,4\right)$

is shown below:

  

Case 2: Consider two sides of parallelogram as $\overline{PR}$ and $\overline{RQ}$ .

The line $\overline{RQ}$ corresponds to the line $x=2$ . So, the side parallel to $\overline{RQ}$ will be $x=-1$ .

Find the line parallel to $\overline{PR}$ and passes through point $Q$ . The slope of line $\overline{PR}$ is:

  $\begin{array}{c}{m}_{PR}=\frac{3-1}{2-\left(-1\right)}\\ =\frac{2}{3}\end{array}$

The equation of line parallel to $\overline{PR}$ is:

  $\begin{array}{c}y=m\left(x-x_1\right)+y_1\\ y-0=\frac{2}{3}\left(x-2\right)\\ y=\frac{2}{3}\left(x-2\right)\end{array}$

To find the fourth vertex, substitute $-1$ for $x$ in the above equation.

  $\begin{array}{l}y=\frac{2}{3}\left(-1-2\right)\\ y=-2\end{array}$

So, the fourth vertex of the parallelogram is $\left(-1,-2\right)$ . The parallelogram with vertices $\left(-1,1\right)$ , $\left(2,0\right)$ , $\left(2,3\right)$ and $\left(-1,-2\right)$

is shown below:

  

Case 2: Consider two sides of parallelogram as $\overline{RP}$ and $\overline{PQ}$ .

First fine the equation of line that is parallel to $\overline{RP}$ and passes through the point $Q$ . The slope of line $\overline{RP}$ is:

  $m_{RP}=\frac{2}{3}$

The equation of line parallel to $\overline{RP}$ is:

  $\begin{array}{c}y=m\left(x-x_1\right)+y_1\\ y-0=\frac{2}{3}\left(x-2\right)\\ y=\frac{2}{3}\left(x-2\right)\end{array}$

First fine the equation of line that is parallel to $\overline{PQ}$ and passes through the point $R$ . The slope of line $\overline{PQ}$ is:

  $m_{PQ}=-\frac{1}{3}$

The equation of line parallel to $\overline{PQ}$ is:

  $y=-\frac{1}{3}\left(x-2\right)+3$

The fourth vertex of the parallelogram is the intersection point of both lines. So,

  $\begin{array}{c}\frac{2}{3}\left(x-2\right)=-\frac{1}{3}\left(x-2\right)+3\\ 2\left(x-2\right)=-x+2+9\\ 2x-4=-x+11\\ x=5\end{array}$

Substitute 5 for $x$ in the equation $y=-\frac{1}{3}\left(x-2\right)+3$ .

  $\begin{array}{c}y=-\frac{1}{3}\left(5-2\right)+3\\ =-1+3\\ =2\end{array}$

So, the fourth vertex of the parallelogram is $\left(5,2\right)$ . The parallelogram with vertices $\left(-1,1\right)$ , $\left(2,0\right)$ , $\left(2,3\right)$ and $\left(5,2\right)$

is shown below:

  

Therefore, the missing vertices of three different parallelograms with three common vertices are $\left(5,2\right)$ , $\left(-1,4\right)$ and $\left(-1,-2\right)$ .