Chapter 11.8, Problem 17HP

To determine

To find: Using at least one dilation, describe a series of transformations where the image is Congruent to preimage

Explanation of Solution

Let us assume a polygon of vertices

  $\begin{array}{l}A=\left(0,5\right)\\ B=\left(0,0\right)\\ C=\left(10,5\right)\end{array}$

First, we will do dilation using scale factor $k=2$

First dilation is $\left(x,y\right)\to \left(2x,2y\right)$ .

Multiply the coordinates of each vertex by $2$ .

  $\begin{array}{l}A\left(0,5\right)\to A\text{'}\left(2\left(0\right),2\left(5\right)\right)\\ \Rightarrow A\left(0,5\right)\to A\text{'}\left(0,10\right)\end{array}$

  $\begin{array}{l}B\left(0,0\right)\to B\text{'}\left(2\left(0\right),2\left(0\right)\right)\\ \Rightarrow B\left(0,0\right)\to B\text{'}\left(0,0\right)\end{array}$

  $\begin{array}{l}C\left(10,5\right)\to C\text{'}\left(2\left(10\right),2\left(5\right)\right)\\ \Rightarrow C\left(10,5\right)\to C\text{'}\left(20,10\right)\end{array}$

Therefore,

Coordinates after dilation are

  $\begin{array}{l}A\text{'}\left(0,10\right)\\ B\text{'}\left(0,0\right)\\ C\text{'}\left(20,10\right)\end{array}$

Now we are doing reflection about Y-axis

If we reflect about Y-axis the x-coordinate of each vertex will multiply by $-1$

Vertices after reflection are

  $\begin{array}{l}A\text{'}\left(0,10\right)\to A\text{'}\text{'}\left(-0,10\right)\\ \Rightarrow A\text{'}\left(0,10\right)\to A\text{'}\text{'}\left(0,10\right)\\ B\text{'}\left(0,0\right)\to B\text{'}\text{'}\left(-0,0\right)\\ \Rightarrow B\text{'}\left(0,0\right)\to B\text{'}\text{'}\left(0,0\right)\\ C\text{'}\left(20,10\right)\to C\text{'}\text{'}(-20,10)\end{array}$

Vertices after reflection are

  $\begin{array}{l}A\text{'}\text{'}\left(0,10\right)\\ B\text{'}\text{'}\left(0,0\right)\\ C\text{'}\text{'}(-20,10)\end{array}$

Again, doing Dilation with scale factor $k=\frac{1}{2}$

We get

Vertices after second dilation are

  $\begin{array}{l}A\text{'}\text{'}\left(0,10\right)\to A\text{'}\text{'}\text{'}\left(\frac{1}{2}\left(0\right),\frac{1}{2}\left(10\right)\right)\\ \Rightarrow A\text{'}\text{'}\left(0,10\right)\to A\text{'}\text{'}\text{'}\left(0,5\right)\end{array}$

  $\begin{array}{l}B\text{'}\text{'}\left(0,0\right)\to B\text{'}\text{'}\text{'}\left(\frac{1}{2}\left(0\right),\frac{1}{2}\left(0\right)\right)\\ \Rightarrow B\text{'}\text{'}\left(0,0\right)\to B\text{'}\text{'}\text{'}\left(0,0\right)\end{array}$

  $\begin{array}{l}C\text{'}\text{'}\left(-20,10\right)\to C\text{'}\text{'}\text{'}\left(\frac{1}{2}\left(-20\right),\frac{1}{2}\left(10\right)\right)\\ \Rightarrow C\text{'}\text{'}\left(-20,10\right)\to C\text{'}\text{'}\text{'}\left(-10,5\right)\end{array}$

Therefore,

Final Coordinates are

  $\begin{array}{l}A\text{'}\text{'}\text{'}\left(0,5\right)\\ B\text{'}\text{'}\text{'}\left(0.0\right)\\ C\text{'}\text{'}\text{'}\left(-10,5\right)\end{array}$

Graph of above situation is

  

From Graph

We can say that

  $\begin{array}{l}AB=5\\ AC=10\end{array}$

  $\begin{array}{l}A\text{'}\text{'}\text{'}B\text{'}\text{'}\text{'}=5\\ A\text{'}\text{'}\text{'}C\text{'}\text{'}\text{'}=10\end{array}$

As $\frac{A\text{'}\text{'}\text{'}B\text{'}\text{'}\text{'}}{AB}=\frac{A\text{'}\text{'}\text{'}C\text{'}\text{'}\text{'}}{AC}=1$

We can say that Both $\Delta ABC$ and $\Delta A\text{'}\text{'}\text{'}B\text{'}\text{'}\text{'}C\text{'}\text{'}\text{'}$ are Congruent to each other.