Chapter 11.8, Problem 16IP

a.

To determine

To find: Vertices of each figure after translation and dilation.

Answer to Problem 16IP

  $\begin{array}{l}A=R\text{'}\text{'}\left(2,2\right)\\ B=S\text{'}\text{'}\left(8,2\right)\\ C=T\text{'}\text{'}\left(6,-2\right)\\ D=U\text{'}\text{'}\left(4,-2\right)\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}R=\left(1,3\right)\\ S=\left(4,3\right)\\ T=\left(3,1\right)\\ U=\left(2,1\right)\end{array}$

First Translation of $2$ units down followed by dilation with a scale factor of $k=2$

Translation of $2$ down mean each vertex is moved $2$ units down along y-axis.

It means we need subtract $2$ from y-coordinate of each vertex.

New vertices after Translation are

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

New vertices after translation are

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

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

Multiply the coordinates of each vertex by $2$ .

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

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

  $\begin{array}{l}T\text{'}\left(3,-1\right)\to T\text{'}\text{'}\left(2\left(3\right),2\left(-1\right)\right)\\ \Rightarrow T\text{'}\left(3,-1\right)\to T\text{'}\text{'}\left(6,-2\right)\end{array}$

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

Therefore,

Coordinates after translation followed by dilation are

  $\begin{array}{l}A=R\text{'}\text{'}\left(2,2\right)\\ B=S\text{'}\text{'}\left(8,2\right)\\ C=T\text{'}\text{'}\left(6,-2\right)\\ D=U\text{'}\text{'}\left(4,-2\right)\end{array}$

Graph of the image is

  

b.

To determine

To find: Vertices of each figure after dilation and translation.

Answer to Problem 16IP

  $\begin{array}{l}E=R\text{'}\text{'}=\left(2,4\right)\\ F=S\text{'}\text{'}=\left(8,4\right)\\ G=T\text{'}\text{'}=\left(6,0\right)\\ H=U\text{'}\text{'}=\left(4,0\right)\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}R=\left(1,3\right)\\ S=\left(4,3\right)\\ T=\left(3,1\right)\\ U=\left(2,1\right)\end{array}$

First Dilation with a scale factor of $k=2$ followed by Translation of $2$ units down

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

Multiply the coordinates of each vertex by $2$ .

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

  $\begin{array}{l}S\left(4,3\right)\to S\text{'}\left(2\left(4\right),2\left(3\right)\right)\\ \Rightarrow S\left(4,3\right)\to S\text{'}\left(8,6\right)\end{array}$

  $\begin{array}{l}T\left(3,1\right)\to T\text{'}\left(2\left(3\right),2\left(1\right)\right)\\ \Rightarrow T\left(3,1\right)\to T\text{'}\left(6,2\right)\end{array}$

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

Vertices after Dilation are

  $\begin{array}{l}R\text{'}=\left(2,6\right)\\ S\text{'}=\left(8,6\right)\\ T\text{'}=\left(6,2\right)\\ U\text{'}=\left(4,2\right)\end{array}$

Translation of $2$ down mean each vertex is moved $2$ units down along y-axis.

It means we need subtract $2$ from y-coordinate of each vertex.

New vertices after Translation are

  $\begin{array}{l}R\text{'}=\left(2,6\right)\\ \Rightarrow R\text{'}\text{'}=\left(2,6-2\right)\\ \Rightarrow R\text{'}\text{'}=\left(2,4\right)\\ S\text{'}=\left(8,6\right)\\ \Rightarrow S\text{'}\text{'}=\left(8,6-2\right)\\ \Rightarrow S\text{'}\text{'}=\left(8,4\right)\\ T\text{'}=\left(6,2\right)\\ \Rightarrow T\text{'}\text{'}=\left(6,2-2\right)\\ \Rightarrow T\text{'}\text{'}=\left(6,0\right)\\ U\text{'}=\left(4,2\right)\\ \Rightarrow U\text{'}\text{'}=\left(4,2-2\right)\\ \Rightarrow U\text{'}\text{'}=\left(4,0\right)\end{array}$

New vertices after dilation and translation are

  $\begin{array}{l}E=R\text{'}\text{'}=\left(2,4\right)\\ F=S\text{'}\text{'}=\left(8,4\right)\\ G=T\text{'}\text{'}=\left(6,0\right)\\ H=U\text{'}\text{'}=\left(4,0\right)\end{array}$

Graph of the image is

  

c.

To determine

To find:Which figures are similar and which figures are Congruent.

Answer to Problem 16IP

The Trapezoid before translation and dilation in Sub-part(a) and The Trapezoid before dilation and translation in Sub-part (b) are Congruent.

Trapezoid $RSTU$ and the Trapezoid $ABCD$ are Similar

Trapezoid $RSTU$ and the Trapezoid $EFGH$ are Similar.

Explanation of Solution

From Answers of Sub-part (a) and Sub-part(b)

We can say that

The Trapezoid before translation and dilation in Sub-part(a) and The Trapezoid before dilation and translation in Sub-part (b) are Congruent.

The Trapezoid $RSTU$ and the Trapezoid $ABCD$ in Sub-part (a) are Similar to each other because as Dilation Occurs here with a scale factor of $k=2$ .

The Trapezoid $RSTU$ and the Trapezoid $EFGH$ in Sub-part (b) are Similar to each other because as Dilation Occurs here with a scale factor of $k=2$ .

d.

To determine

To find:What transformation would map $ABCD$ onto $EFGH$

Answer to Problem 16IP

Trapezoid $ABCD$ is Translated $2$ units Up to get Trapezoid $EFGH$

Explanation of Solution

From above Sub-parts(a) and Sub-part(b)

We get

Vertices of Trapezoid $RSTU$ are

  $\begin{array}{l}R=\left(1,3\right)\\ S=\left(4,3\right)\\ T=\left(3,1\right)\\ U=\left(2,1\right)\end{array}$

Vertices of Trapezoid $EFGH$ are

  $\begin{array}{l}E=\left(2,4\right)\\ F=\left(8,4\right)\\ G=\left(6,0\right)\\ H=\left(4,0\right)\end{array}$

Vertices of Trapezoid $ABCD$ are

  $\begin{array}{l}A=R\text{'}\text{'}\left(2,2\right)\\ B=S\text{'}\text{'}\left(8,2\right)\\ C=T\text{'}\text{'}\left(6,-2\right)\\ D=U\text{'}\text{'}\left(4,-2\right)\end{array}$

From above information

We can say that

Trapezoid $ABCD$ is Translated $2$ units Up to get Trapezoid $EFGH$