Chapter 7.6, Problem 3BCYP

To determine

Draw a graph for the original figure and its dilated image. Check whether the dilation is a similarity transformation or not.

Answer to Problem 3BCYP

The dilation is a similarity transformation.

Explanation of Solution

Given:

The given vertices are

original $\to H\left(0,0\right),J\left(6,0\right),K\left(6,4\right),L\left(0,4\right)$

image $\to W\left(0,0\right),X\left(3,0\right),Y\left(3,2\right),Z\left(0,2\right)$

Calculation:

The graph for the points is given below.

  

Find the distance (using distance formula) of corresponding sides and find the ratio.

  $\begin{array}{l}WZ=\sqrt{{\left(0-0\right)}^2+{\left(0-2\right)}^2}=\sqrt{4}=2\\ HL=\sqrt{{\left(0-0\right)}^2+{\left(0-4\right)}^2}=\sqrt{16}=4\end{array}$

  $\frac{WZ}{HL}=\frac{2}{4}=\frac{1}{2}$

  $\begin{array}{l}WX=\sqrt{{\left(0-3\right)}^2+{\left(0-0\right)}^2}=\sqrt{9}=3\\ HJ=\sqrt{{\left(0-6\right)}^2+{\left(0-0\right)}^2}=\sqrt{36}=6\end{array}$

  $\frac{WX}{HJ}=\frac{3}{6}=\frac{1}{2}$

  $\begin{array}{l}ZY=\sqrt{{\left(0-3\right)}^2+{\left(2-2\right)}^2}=\sqrt{9}=3\\ LK=\sqrt{{\left(0-6\right)}^2+{\left(4-4\right)}^2}=\sqrt{36}=6\end{array}$

  $\frac{ZY}{LK}=\frac{3}{6}=\frac{1}{2}$

  $\begin{array}{l}XY=\sqrt{{\left(3-3\right)}^2+{\left(0-2\right)}^2}=\sqrt{4}=2\\ JK=\sqrt{{\left(6-6\right)}^2+{\left(0-4\right)}^2}=\sqrt{16}=4\end{array}$

  $\frac{XY}{JK}=\frac{3}{6}=\frac{1}{2}$

From the above result, the length of the sides are proportional.

Hence the dilation is a similarity transformation.