Chapter 11.7, Problem 13IP

To determine

To find: Vertices of each figure after a dilation.

Answer to Problem 13IP

  $\begin{array}{l}A\text{'}\left(\frac{-3}{2},\frac{-5}{2}\right)\\ B\text{'}\left(0,1\right)\\ C\text{'}\left(\frac{3}{2},\frac{-1}{2}\right)\\ D\text{'}\left(0,-2\right)\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}k=\frac{1}{2}\\ A=\left(-3,-5\right)\\ B=\left(0,2\right)\\ C=\left(3,-1\right)\\ D=\left(0,-4\right)\end{array}$

The dilation is $\left(x,y\right)\to \left(\frac{1}{2}x,\frac{1}{2}y\right)$ .

Multiply the coordinates of each vertex by $\frac{1}{2}$ .

  $\begin{array}{l}A\left(-3,-5\right)A\text{'}\left(\frac{1}{2}\left(-3\right),\frac{1}{2}\left(-5\right)\right)\\ \Rightarrow A\left(-3,-5\right)\to A\text{'}\left(\frac{-3}{2},\frac{-5}{2}\right)\end{array}$

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

  $\begin{array}{l}C\left(3,-1\right)\to C\text{'}\left(\frac{1}{2}\left(3\right),\frac{1}{2}\left(-1\right)\right)\\ \Rightarrow C\left(3,-1\right)\to C\text{'}\left(\frac{3}{2},\frac{-1}{2}\right)\end{array}$

  $\begin{array}{l}D\left(0,-4\right)\to D\text{'}\left(\frac{1}{2}\left(0\right),\frac{1}{2}\left(-4\right)\right)\\ \Rightarrow D\left(0,-4\right)\to D\text{'}\left(0,-2\right)\end{array}$

Therefore,

Coordinates after dilation are

  $\begin{array}{l}A\text{'}\left(\frac{-3}{2},\frac{-5}{2}\right)\\ B\text{'}\left(0,1\right)\\ C\text{'}\left(\frac{3}{2},\frac{-1}{2}\right)\\ D\text{'}\left(0,-2\right)\end{array}$

Graph of image is