Question

In Exercise 18-20, refer to the diagrams om page 578. Given the reflection $R_m:\overrightarrow{PQ}\to \overrightarrow{P\text{'}Q\text{'}}$, write the key steps of a proof that $PQ=P\text{'}Q\text{'}$ for each case.

Case 4.

Short answer

It is proved that, $PQ=P\text{'}Q\text{'}$.

Step-by-step solution

Step 1. Given Information.

It is given that$\overline{P\text{'}Q\text{'}}$is reflection of PQ in line m.

Step 2. Proof.

Since, $\overline{P\text{'}Q\text{'}}$is reflection of PQin line m and it is visible in the figure that P and Q are not on the line m hence, m is perpendicular bisector of $PP\text{'}\text{and}QQ\text{'}$.

And if m is perpendicular bisector of both $PP\text{'}\text{and}QQ\text{'}$. Since, $PP\text{'}\text{and}QQ\text{'}$are not on same line then it obvious that$PP\text{'}\text{and}QQ\text{'}$are parallel.

The two statements that m is perpendicular bisector of $PP\text{'}\text{and}QQ\text{'}$,$PP\text{'}$ is parallel to$QQ\text{'}$leads to conclude that $PQ=P\text{'}Q\text{'}$.

Step 3. Conclusion.

It is proved that, $PQ=P\text{'}Q\text{'}$.