Chapter 14.6, Problem 25WE

(a)

To determine

To plot: The point $P\left(5,2\right)$ and its image $Q$ under the mapping $R_y\circ R_l$ .

Explanation of Solution

Given information: The reflection in the line $y=x$ be $R_l$ and the reflection in y -axis be $R_y$ . The given point is $\left(5,2\right)$ .

Graph:

If the reflection line is $y=x$ , then the mapping is:

  $R_l:\left(x_1,y_1\right)\to \left(y_1,x_1\right)$

If the reflection line is y -axis, then the mapping is:

  $R_y:\left(x_1,y_1\right)\to \left(-x_1,y_1\right)$

The composition of $R_l$ and $R_y$ is:

  $R_y\circ R_l:\left(x_1,y_1\right)\to \left(-y_1,x_1\right)$

So, the image of point $P\left(5,2\right)$ under the mapping $R_y\circ R_l$ is:

  $R_y\circ R_l:\left(5,2\right)\to \left(-2,5\right)$

Plot the point $P\left(5,2\right)$ and its image $Q\left(-2,5\right)$ on the coordinate plane as shown below.

  

Figure (1)

Interpretation: From the figure it can be observed that to draw the image of point Q , plot the image of point $P\left(5,2\right)$ reflection in the line $y=x$ and then plot its image in the reflection of y -axis.

(b)

To determine

To find: The value of $m\angle POQ$ .

Answer to Problem 25WE

The value of $m\angle POQ$ is $90°$ .

Explanation of Solution

Given information: The reflection in the line $y=x$ be $R_l$ and the reflection in y -axis be $R_y$ .

Calculation:

As per the theorem 14-8, composite of reflections in two intersecting lines is a rotation about the point of intersection of the two lines. The angle of rotation is twice the angle from the first line of reflection to the second.

The angle of rotation or $m\angle POQ$ is twice the angle from first line of reflection to the second line. The angle between $y=x$ and y -axis is 45 degree. So,

  $\begin{array}{c}m\angle POQ=2\times 45°\\ =90°\end{array}$

Therefore, the value of $m\angle POQ$ is $90°$ .

(c)

To determine

To verify: The line $\overline{OP}$ is perpendicular to the line $\overline{OQ}$ by using the slopes of these lines.

Explanation of Solution

Given information: The reflection in the line $y=x$ be $R_l$ and the reflection in y -axis be $R_y$ . The given point is $\left(5,2\right)$ .

Proof:

From part (a), the image of point $P\left(5,2\right)$ is $Q\left(-2,5\right)$ . The coordinates of origin O is $\left(0,0\right)$ .

The slope of line OP that passes through the points $O\left(0,0\right)$ and $P\left(5,2\right)$ is:

  $\begin{array}{c}{m}_1=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{2-0}{5-0}\\ =\frac{2}{5}\end{array}$

The slope of line OQ that passes through the points $O\left(0,0\right)$ and $Q\left(-2,5\right)$ is:

  $\begin{array}{c}{m}_2=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{5-0}{-2-0}\\ =\frac{5}{-2}\end{array}$

Now,

  $\begin{array}{c}{m}_1{m}_2=\frac{2}{5}\times \frac{5}{-2}\\ =-1\end{array}$

As the product of lines OP and OQ is $-1$ . So, both lines are perpendicular to each other.

Hence, it is proved that $\overline{OP}\perp \overline{OQ}$ .

(d)

To determine

To find: The image of $\left(x,y\right)$ under the mappings $R_y\circ R_l$ and $R_l\circ R_y$ .

Answer to Problem 25WE

The image of $\left(x,y\right)$ under the mapping $R_y\circ R_l$ is $\left(-y,x\right)$ and $R_l\circ R_y$ is $\left(y,-x\right)$ .

Explanation of Solution

Given information: The reflection in the line $y=x$ be $R_l$ and the reflection in y -axis be $R_y$ .

Calculation:

The image of point $\left(x,y\right)$ under the reflection $y=x$ is:

  $R_l:\left(x,y\right)\to \left(y,x\right)$

The image of point $\left(x,y\right)$ along the reflection y -axis is:

  $R_y:\left(x,y\right)\to \left(-x,y\right)$

Now, the image of point $\left(x,y\right)$ under the composition of $R_y$ and $R_l$ is:

  $R_y\circ R_l:\left(x,y\right)\to \left(-y,x\right)$

Now, the image of point $\left(x,y\right)$ under the composition of $R_l$ and $R_y$ is:

  $R_l\circ R_y:\left(x,y\right)\to \left(y,-x\right)$

Therefore, the image of $\left(x,y\right)$ under the mapping $R_y\circ R_l$ is $\left(-y,x\right)$ and $R_l\circ R_y$ is $\left(y,-x\right)$ .