Chapter 9.4, Problem 25PPS

To determine

To find: The graph and its mirror image.

Answer to Problem 25PPS

The graph and its mirror image is shown in Figure 1.

Explanation of Solution

Given:

Consider the given diagram.

  

Calculation:

For origin,

  $\begin{array}{l}\left(x,y\right)\to \left(-y,x\right)\\ A\left(0,1\right)\to A\text{'}\left(-1,0\right)\\ B\left(-2,-5\right)\to B\text{'}\left(5,-2\right)\end{array}$

Calculate the slope.

  $\begin{array}{c}m=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{-2-0}{5-\left(-1\right)}\\ =-\frac{1}{3}\end{array}$

Write the equation of line.

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-0=-\frac{1}{3}\left(x+1\right)\\ y=-\frac{1}{3}x-\frac{1}{3}\end{array}$

For x-axis,

  $\begin{array}{l}\left(x,y\right)\to \left(-y,x\right)\\ A\text{'}\left(-1,0\right)\to A\text{'}\text{'}\left(-1,0\right)\\ B\text{'}\left(5,-2\right)\to B\text{'}\text{'}\left(5,2\right)\end{array}$

Calculate the slope.

  $\begin{array}{c}m=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{2-0}{5-\left(-1\right)}\\ =\frac{1}{3}\end{array}$

Write the equation of line.

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-0=\frac{1}{3}\left(x+1\right)\\ y=\frac{1}{3}x+\frac{1}{3}\end{array}$

The required diagram is shown below:

  

Figure 1

Thus, the graph and its mirror image is shown in Figure 1.