Chapter 7.3, Problem 32WE

To determine

To Plot: The graph for given points of the quadrilateral.

Explanation of Solution

Given information:

The quadrilateral $ABCD$ with vertices $A\left(0,0\right),B\left(4,0\right),C\left(2,4\right),D\left(0,2\right)$ .

The vertices of the similar quadrilateral are $A\text{'}\left(-10,-2\right),B\text{'}\left(-2,-2\right)$ .

The ray $\overline{A\text{'}B\text{'}}$ is equal to $\overline{AB}$ .

Therefore, the point is $D^{\prime }(-10,2)$ .

The slope of the line connecting vertices $C$ and $D$ is $\frac{4-2}{2-0}=1$

Slope of the line connecting $C^{\prime }$ and $D^{\prime }$ is 1 .

The equation of the segment

  $\overline{C^{\prime }{D}^{\prime }}$ is $y=x+c_1$ since this line passes through (-10,2)

The equation becomes;

  $\begin{array}{l}2=-10+c_1\\ c_1=12\end{array}$

Thus, the equation of line $\overline{C^{\prime }{D}^{\prime }}$ is;

  $y=x+12$

The slope of the line connecting vertices $C$ and $B$ is $\frac{4-0}{2-4}=-2$ .

Slope of the line connecting $C^{\prime }$ and $B^{\prime }$ is $1.$

The equation of the segment $\overline{C^{\prime }{B}^{\prime }}$ is $y=-2x+c_2$ since this line passes through (-2,-2) ,

And the equation is;

  $\begin{array}{l}-2=-2\times (-2)+c_2\hfill \\ c_2=-6\hfill \end{array}$

Thus, the equation of line $\overline{C^{\prime }{B}^{\prime }}$ is $y=-2x-6$

Solve the equations, $y=x+12$ and $y=-2x-6,$ the vertices of $D^{\prime }$ can be obtained.

  $\begin{array}{l}-2x-6=x+12\\ \begin{array}{l}x=-6\hfill \\ y=6\hfill \end{array}\end{array}$

Therefore the point is $C^{\prime }=(-6,6)$ .

Connect the points $A^{\prime }(-10,-2),B^{\prime }(-2,-2),C^{\prime }(-6,6)$ and $D^{\prime }(-10,2)$ the quadrilateral

  $A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }$ similar to $ABCD$ is obtained.

The quadrilaterals are plotted below using Maple.