Chapter 14.6, Problem 26WE

(a)

To determine

To plot: The point $P\left(-6,-2\right)$ and its image $Q$ under the mapping $R_k\circ R_x$ .

Explanation of Solution

Given information: The reflection in the line $y=-x$ be $R_k$ and the reflection in x -axis be $R_x$ . The given point is $\left(-6,-2\right)$ .

Graph:

The image of point $\left(-6,-2\right)$ under the reflection line is $y=-x$ is:

  $R_k:\left(-6,-2\right)\to \left(2,6\right)$

The image of point $\left(2,6\right)$ under the reflection line is x -axis is:

  $R_x:\left(2,6\right)\to \left(2,-6\right)$

So, the image of point $P\left(-6,-2\right)$ under the mapping $R_k\circ R_x$ is:

  $R_k\circ R_x:\left(-6,-2\right)\to \left(2,-6\right)$

Plot the point $P\left(-6,-2\right)$ and its image $Q\left(2,-6\right)$ 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 $P\left(-6,-2\right)$ reflection in the line $y=-x$ and then plot its image in the reflection of x -axis.

(b)

To determine

To show: the measure of $\angle POQ$ is $90°$ .

Explanation of Solution

Proof:

From part (a), the image of point $P\left(-6,-2\right)$ is $Q\left(2,-6\right)$ . 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 $P\left(-6,-2\right)$ is:

  $\begin{array}{c}{m}_1=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{-2-0}{-6-0}\\ =\frac{2}{6}\\ =\frac{1}{3}\end{array}$

The slope of line OQ that passes through the points $O\left(0,0\right)$ and $Q\left(2,-6\right)$ is:

  $\begin{array}{c}{m}_2=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{-6-0}{2-0}\\ =-3\end{array}$

It is known that the condition for two perpendicular lines is $m_1{m}_2=-1$ .

Now, the product of slope of lines OP and OQ is:

  $\begin{array}{c}{m}_1{m}_2=\frac{1}{3}\times -3\\ =-1\end{array}$

So, both lines are perpendicular to each other. The angle between two perpendicular lines is $90°$ .

Hence, it is proved that $m\angle POQ=90°$ .

(c)

To determine

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

Answer to Problem 26WE

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 line is $y=-x$ is:

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

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

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

So, the image of point $\left(x,y\right)$ under the mapping $R_k\circ R_x$ is:

  $R_k\circ R_x:\left(x,y\right)\to \left(-y,x\right)$

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

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

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

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

So, the image of point $\left(x,y\right)$ under the mapping $R_x\circ R_k$ is:

  $R_x\circ R_k:\left(x,-y\right)\to \left(y,-x\right)$

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 $\left(y,-x\right)$ .