Chapter 13.5, Problem 15PSC

(a)

To determine

To find:The area of the reflected square.

Answer to Problem 15PSC

The area of the reflected square is 10.

Explanation of Solution

Given:

The vertices of the square is,

  $\begin{array}{l}A=\left(1,1\right)\\ B=\left(0,4\right)\\ C=\left(3,5\right)\\ D=\left(4,2\right)\end{array}$

Calculation:

Calculate the side of the square.

  $\begin{array}{c}a=\sqrt{{\left(1-0\right)}^2+{\left(-1+4\right)}^2}\\ =\sqrt{1+9}\\ =\sqrt{10}\end{array}$

Calculate the area.

  $\begin{array}{c}\text{Area}=a^2\\ ={\left(\sqrt{10}\right)}^2\\ =10\end{array}$

Thus, the area of the reflected square is 10.

(b)

To determine

To find:The equation of $A\text{'}C\text{'}$ .

Answer to Problem 15PSC

The equation of $A\text{'}C\text{'}$ is $y=-2x+1$ .

Explanation of Solution

Given:

The vertices of the square is,

  $\begin{array}{l}A=\left(1,1\right)\\ B=\left(0,4\right)\\ C=\left(3,5\right)\\ D=\left(4,2\right)\end{array}$

Calculation:

Calculate the slope.

  $\begin{array}{c}m=\frac{-5+1}{3-1}\\ =-2\end{array}$

Calculate the equation of line.

  $\begin{array}{l}y=mx+c\\ -1=-2\left(1\right)+b\\ b=1\end{array}$

Now,

  $\begin{array}{l}y=mx+c\\ y=-2x+1\end{array}$

Thus, the equation of $A\text{'}C\text{'}$ is $y=-2x+1$ .