Chapter 13, Problem 36RP

(a)

To determine

To find: The image of the point $\left(-5,10\right)$ when the image is reflected over the x axis.

Answer to Problem 36RP

The reflected image is $\left(-5,-10\right)$ .

Explanation of Solution

When image is reflected over the x axis the coordinates of the x axis are same but the y axis is reversed.

Then, the reflected image is,

  $\left(-5,-10\right)$

(b)

To determine

To find: The image of the point $\left(-5,10\right)$ when the image is reflected over $\left(-3,1\right)$ .

Answer to Problem 36RP

The reflected image is $\left(-1,-8\right)$ .

Explanation of Solution

Consider the point $\left(-3,1\right)$ is the mid-point of the point $\left(-5,10\right)$ and $\left(x,y\right)$ then,

  $\left(-3,1\right)=\left(\frac{-5+x}{2},\frac{10+y}{2}\right)$

Then,

  $\begin{array}{l}\frac{-5+x}{2}=-3\\ x=-1\end{array}$

Also,

  $\begin{array}{l}\frac{10+y}{2}=1\\ y=-8\end{array}$

Thus, the reflected image is $\left(-1,-8\right)$ .

(c)

To determine

To find: The image of the point $\left(-5,10\right)$ when the image is reflected over graph of $y=2x$ .

Answer to Problem 36RP

The reflected image is $\left(11,2\right)$ .

Explanation of Solution

Consider the slope of the perpendicular is $\frac{-1}{2}$ then the line crosses the line for the point $\left(-5,10\right)$ at the points $\left(3,6\right)$ . Thus, $\left(3,6\right)$ is the mid-point $\left(-5,10\right)$ and $\left(x,y\right)$ . Then,

  $\left(3,6\right)=\left(\frac{-5+x}{2},\frac{10+y}{2}\right)$

Then,

  $\begin{array}{l}\frac{-5+x}{2}=3\\ x=11\end{array}$

Also,

  $\begin{array}{l}\frac{10+y}{2}=6\\ y=2\end{array}$

Thus, the reflected image is $\left(11,2\right)$ .