Chapter 11.7, Problem 9IP

To determine

To find: Vertices of each figure after a dilation.

Answer to Problem 9IP

  $\begin{array}{l}N\text{'}\left(0,\frac{5}{3}\right)\\ K\text{'}\left(\frac{5}{3},0\right)\\ L\text{'}\left(\frac{8}{3},1\right)\\ M\text{'}\left(1,\frac{8}{3}\right)\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}k=\frac{1}{3}\\ N=\left(0,5\right)\\ K=\left(5,0\right)\\ L=\left(8,3\right)\\ M=\left(3,8\right)\end{array}$

The dilation is $\left(x,y\right)\to \left(\frac{1}{3}x,\frac{1}{3}y\right)$ .

Multiply the coordinates of each vertex by $\frac{1}{3}$ .

  $\begin{array}{l}N\left(0,5\right)\to N\text{'}\left(\frac{1}{3}\left(0\right),\frac{1}{3}\left(5\right)\right)\\ \Rightarrow N\left(0,5\right)\to N\text{'}\left(0,\frac{5}{3}\right)\end{array}$

  $\begin{array}{l}K\left(5,0\right)\to K\text{'}\left(\frac{1}{3}\left(5\right),\frac{1}{3}\left(0\right)\right)\\ \Rightarrow K\left(5,0\right)\to K\text{'}\left(\frac{5}{3},0\right)\end{array}$

  $\begin{array}{l}L\left(8,3\right)\to L\text{'}\left(\frac{1}{3}\left(8\right),\frac{1}{3}\left(3\right)\right)\\ \Rightarrow L\left(8,3\right)\to L\text{'}\left(\frac{8}{3},1\right)\end{array}$

  $\begin{array}{l}M\left(3,8\right)\to M\text{'}\left(\frac{1}{3}\left(3\right),\frac{1}{3}\left(8\right)\right)\\ \Rightarrow M\left(3,8\right)\to M\text{'}\left(1,\frac{8}{3}\right)\end{array}$

Therefore,

Coordinates after dilation are

  $\begin{array}{l}N\text{'}\left(0,\frac{5}{3}\right)\\ K\text{'}\left(\frac{5}{3},0\right)\\ L\text{'}\left(\frac{8}{3},1\right)\\ M\text{'}\left(1,\frac{8}{3}\right)\end{array}$

Graph of image is