Chapter 11.7, Problem 2GP

To determine

To find: Vertices of each figure after a dilation.

Answer to Problem 2GP

  $\begin{array}{l}R\text{'}\left(\frac{1}{4},\frac{3}{4}\right)\\ S\text{'}\left(2,\frac{1}{2}\right)\\ U\text{'}\left(\frac{3}{2},\frac{3}{2}\right)\\ T\text{'}\left(\frac{5}{4},-\frac{1}{4}\right)\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}k=\frac{1}{4}\\ R=\left(1,3\right)\\ S=\left(8,2\right)\\ U=\left(6,6\right)\\ T=\left(5,-1\right)\end{array}$

The dilation is $\left(x,y\right)\to \left(\frac{1}{4}x,\frac{1}{4}y\right)$ .

Multiply the coordinates of each vertex by $\frac{1}{4}$ .

  $\begin{array}{l}R\left(1,3\right)\to R\text{'}\left(\frac{1}{4}\left(1\right),\frac{1}{4}\left(3\right)\right)\\ \Rightarrow R\left(1,3\right)\to R\text{'}\left(\frac{1}{4},\frac{3}{4}\right)\end{array}$

  $\begin{array}{l}S\left(8,2\right)\to S\text{'}\left(\frac{1}{4}\left(8\right),\frac{1}{4}\left(2\right)\right)\\ \Rightarrow S\left(1,2\right)\to S\text{'}\left(2,\frac{1}{2}\right)\end{array}$

  $\begin{array}{l}U\left(6,6\right)\to U\text{'}\left(\frac{1}{4}\left(6\right),\frac{1}{4}\left(6\right)\right)\\ \Rightarrow U\left(6,6\right)\to U\text{'}\left(\frac{3}{2},\frac{3}{2}\right)\end{array}$

  $\begin{array}{l}T\left(5,-1\right)\to T\text{'}\left(\frac{1}{4}\left(5\right),\frac{1}{4}\left(-1\right)\right)\\ \Rightarrow T\left(5,-1\right)\to T\text{'}\left(\frac{5}{4},\frac{-1}{4}\right)\end{array}$

Therefore,

Coordinates after dilation are

  $\begin{array}{l}R\text{'}\left(\frac{1}{4},\frac{3}{4}\right)\\ S\text{'}\left(2,\frac{1}{2}\right)\\ U\text{'}\left(\frac{3}{2},\frac{3}{2}\right)\\ T\text{'}\left(\frac{5}{4},-\frac{1}{4}\right)\end{array}$

Graph of the image is