Chapter 11.7, Problem 23HP

To determine

To find:Does the order in which you perform multiple dilations always, sometimes, never result in same image.

Answer to Problem 23HP

Order in which you perform multiple dilations always results in same image.

Explanation of Solution

Order in which you perform multiple dilations always results in same image.

Example:

Given

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

  $k=2$ followed by $k=\frac{2}{5}$

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

Multiply the coordinates of each vertex by $2$ .

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

Again, on dilation with scale factor $k=\frac{2}{5}$

We get

  $\begin{array}{l}A\text{'}\left(0,10\right)\to A\text{'}\text{'}\left(\frac{2}{5}\left(0\right),\frac{2}{5}\left(10\right)\right)\\ \Rightarrow A\text{'}\left(0,10\right)\to A\text{'}\text{'}\left(0,4\right)\end{array}$

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

Again, on dilation with scale factor $k=\frac{2}{5}$

We get

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

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

Again, on dilation with scale factor $k=\frac{2}{5}$

We get

  $\begin{array}{l}C\text{'}\left(20,10\right)\to C\text{'}\text{'}\left(\frac{2}{5}\left(20\right),\frac{2}{5}\left(10\right)\right)\\ \Rightarrow C\text{'}\left(20,10\right)\to C\text{'}\text{'}\left(8,4\right)\end{array}$

Therefore,

Coordinates after dilation are

  $\begin{array}{l}A\text{'}\text{'}\left(0,4\right)\\ B\text{'}\text{'}\left(0,0\right)\\ C\text{'}\text{'}\left(8,4\right)\end{array}$

Now,$k=\frac{2}{5}$ followed by $k=2$

Multiply the coordinates of each vertex by $\frac{2}{5}$ .

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

Again, on dilation with scale factor $k=2$

We get

  $\begin{array}{l}A\text{'}\left(0,2\right)\to A\text{'}\text{'}\left(2\left(0\right),2\left(2\right)\right)\\ \Rightarrow A\text{'}\left(0,2\right)\to A\text{'}\text{'}\left(0,4\right)\end{array}$

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

Again, on dilation with scale factor $k=2$

We get

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

  $\begin{array}{l}C\left(10,5\right)\to C\text{'}\left(\frac{2}{5}\left(10\right),\frac{2}{5}\left(5\right)\right)\\ \Rightarrow C\left(10,5\right)\to C\text{'}\left(4,2\right)\end{array}$

Again, on dilation with scale factor $k=2$

We get

  $\begin{array}{l}C\text{'}\left(4,2\right)\to C\text{'}\text{'}\left(2\left(4\right),2\left(2\right)\right)\\ \Rightarrow C\text{'}\left(4,2\right)\to C\text{'}\text{'}\left(8,4\right)\end{array}$

Therefore,

Coordinates after dilation are

  $\begin{array}{l}A\text{'}\text{'}\left(0,4\right)\\ B\text{'}\text{'}\left(0,0\right)\\ C\text{'}\text{'}\left(8,4\right)\end{array}$

On comparing we can say that both images are same.