Chapter 14.1, Problem 23WE

(a)

To determine

To plot: The points $A\left(6,1\right)$ , $B\left(3,4\right)$ , and $C\left(1,-3\right)$ , and their images $A^{\prime }$ , $B^{\prime }$ , and $C^{\prime }$ under the transformation $R:\left(x,y\right)\to \left(-x,y\right)$ .

Explanation of Solution

Given information: The points are $A\left(6,1\right)$ , $B\left(3,4\right)$ , and $C\left(1,-3\right)$ , and the transformation is $R:\left(x,y\right)\to \left(-x,y\right)$ .

Graph:

The images $A^{\prime }$ , $B^{\prime }$ , and $C^{\prime }$ under the transformation $R:\left(x,y\right)\to \left(-x,y\right)$ are as follows:

  $\begin{array}{l}T:A\left(6,1\right)\to A^{\prime }\left(-6,1\right)\\ T:B\left(3,4\right)\to B^{\prime }\left(-3,4\right)\\ T:C\left(1,-3\right)\to C^{\prime }\left(-1,-3\right)\end{array}$

Now, plot all the points and their images on the coordinate plane as shown below.

  

Figure (1)

Interpretation: From the graph it can be observed that the points A , B , C and their images $A^{\prime }$ , $B^{\prime }$ , $C^{\prime }$ are at same distance from the y -axis.

(b)

To determine

To prove: The transformation $R$ is an isometry.

Explanation of Solution

Given information: The points are $A\left(6,1\right)$ , $B\left(3,4\right)$ , and $C\left(1,-3\right)$ . The transformation is $R:\left(x,y\right)\to \left(-x,y\right)$ .

Proof:

Let $P\left(x_1,y_1\right)$ and $Q\left(x_2,y_2\right)$ be any two points. The image of the points under the transformation $R:\left(x,y\right)\to \left(-x,y\right)$ is:

  $\begin{array}{l}R:P\left(x_1,y_1\right)\to P^{\prime }\left(-x_1,y_1\right)\\ R:Q\left(x_2,y_2\right)\to Q^{\prime }\left(-x_2,y_2\right)\end{array}$

To find the distance between two points, use the following distance formula.

  $D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

The distance between points $P$ and $Q$ is:

  $PQ=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

The distance between points $P^{\prime }$ and $Q^{\prime }$ is:

  $\begin{array}{c}{P}^{\prime }{Q}^{\prime }=\sqrt{{\left(\left(-x_2\right)-\left(-x_1\right)\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(-\left[x_2-x_1\right]\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\end{array}$

So, $PQ=P^{\prime }{Q}^{\prime }$ .

Hence, it is proved that the transformation $R$ is an isometry.