Chapter 14.5, Problem 15CE

a.

To determine

To find $OQ\text{'}$ if $OQ=2$ .

Answer to Problem 15CE

The solution is $OQ\text{'}=6$ .

Explanation of Solution

Given information:

The given statement is:

“The dilation $D_{O,3}$ maps P to P’ and Q to Q’.”

Calculation:

The solution can be obtained by:

  $\begin{array}{l}OQ\text{'}=3\cdot OQ=3\cdot 2=6\\ OQ\text{'}=6\end{array}$

Hence, the solution is $OQ\text{'}=6$ .

b.

To determine

To find $P\text{'}Q\text{'}$ if $PQ=7$ .

Answer to Problem 15CE

The solution is $P\text{'}Q\text{'}=21$ .

Explanation of Solution

Given information:

The given statement is:

“The dilation $D_{O,3}$ maps P to P’ and Q to Q’.”

Calculation:

The solution can be obtained by:

  $\begin{array}{l}P\text{'}Q\text{'}=3\cdot PQ=3\cdot 7=21\\ P\text{'}Q\text{'}=21\end{array}$

Hence, the solution is $P\text{'}Q\text{'}=21$ .

c.

To determine

To find $OP$ if $PP\text{'}=10$ .

Answer to Problem 15CE

The solution is $PP\text{'}=5$ .

Explanation of Solution

Given information:

The given statement is:

“The dilation $D_{O,3}$ maps P to P’ and Q to Q’.”

Calculation:

Let OP be $x$ .

The solution can be obtained by:

  $\begin{array}{l}OP\text{'}=3\cdot OP=3\cdot x=3x\\ OP\text{'}=3x\\ PP\text{'}=OP\text{'}-OP\\ PP\text{'}=3x-x=10\\ PP\text{'}=2x=10\\ PP\text{'}=x=5\\ PP\text{'}=5\end{array}$

Hence, the solution is $PP\text{'}=5$ .