Chapter 11, Problem 26CR

To determine

To find: Vertices of each figure after a dilation.

Answer to Problem 26CR

Coordinates after dilation are

  $\begin{array}{l}W\text{'}\left(-2,0\right)\\ X\text{'}\left(\frac{4}{3},4\right)\\ Y\text{'}\left(4,\frac{4}{3}\right)\\ Z\text{'}\left(\frac{4}{3},-\frac{4}{3}\right)\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}k=\frac{2}{3}\\ W=\left(-3,0\right)\\ X=\left(2,6\right)\\ Y=\left(6,2\right)\\ Z=\left(2,-2\right)\end{array}$

The dilation is $\left(x,y\right)\to \left(\frac{2}{3}x,\frac{2}{3}y\right)$ .

Multiply the coordinates of each vertex by $\frac{2}{3}$ .

  $\begin{array}{l}W\left(-3,0\right)\to W\text{'}\left(\frac{2}{3}\left(-3\right),\frac{2}{3}\left(0\right)\right)\\ \Rightarrow W\left(-3,0\right)\to W\text{'}\left(-2,0\right)\end{array}$

  $\begin{array}{l}X\left(2,6\right)\to X\text{'}\left(\frac{2}{3}\left(2\right),\frac{2}{3}\left(6\right)\right)\\ \Rightarrow X\left(2,6\right)\to X\text{'}\left(\frac{4}{3},4\right)\end{array}$

  $\begin{array}{l}Y\left(6,2\right)\to Y\text{'}\left(\frac{2}{3}\left(6\right),\frac{2}{3}\left(2\right)\right)\\ \Rightarrow Y\left(6,2\right)\to Y\text{'}\left(4,\frac{4}{3}\right)\end{array}$

  $\begin{array}{l}Z\left(2,-2\right)\to Z\text{'}\left(\frac{2}{3}\left(2\right),\frac{2}{3}\left(-2\right)\right)\\ \Rightarrow Z\left(2,-2\right)\to Z\text{'}\left(\frac{4}{3},\frac{-4}{3}\right)\end{array}$

Therefore,

Coordinates after dilation are

  $\begin{array}{l}W\text{'}\left(-2,0\right)\\ X\text{'}\left(\frac{4}{3},4\right)\\ Y\text{'}\left(4,\frac{4}{3}\right)\\ Z\text{'}\left(\frac{4}{3},-\frac{4}{3}\right)\end{array}$

Graph of image is