Chapter 14.1, Problem 5CE

a.

To determine

To plot: The given points and their images under the transformation $T\left(x,y\right)=\left(x+1,y+2\right)$ .

Explanation of Solution

Given information: The transformation is $T:\left(x,y\right)\to \left(x+1,y+2\right)$ . The given points are $A\left(0,0\right)$ , $B\left(3,4\right)$ , $C\left(5,1\right)$ and $D\left(-1,-3\right)$ .

Graph:

The transformation of the point $A\left(0,0\right)$ is:

  $T:A\left(0,0\right)\to \left(1,2\right)$

The transformation of the point $B\left(3,4\right)$ is:

  $T:B\left(3,4\right)\to \left(4,6\right)$

The transformation of the point $C\left(5,1\right)$ is:

  $T:C\left(5,1\right)\to \left(6,3\right)$

The transformation of the point $D\left(-1,-3\right)$ is:

  $T:D\left(-1,-3\right)\to \left(0,-1\right)$

The points and images of their transformation is shown below.

  

Figure (1)

Interpretation: From the above figure it can be seen that the transformation is a type of translation of the points.

b.

To determine

To find: The distance $AB$ , $A^{\prime }{B}^{\prime }$ , $CD$ and $C^{\prime }{D}^{\prime }$ .

Answer to Problem 5CE

The length of $AB$ is 5, $A^{\prime }{B}^{\prime }$ is 5, $CD$ is 7.2 and $C^{\prime }{D}^{\prime }$ is 7.2.

Explanation of Solution

Given information: The given points are $A\left(0,0\right)$ , $B\left(3,4\right)$ , $C\left(5,1\right)$ and $D\left(-1,-3\right)$ .

Formula used: The formula to find the distance between two points $\left(x_1,y_1\right)$ and $\left(y_1,y_2\right)$ is:

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

Calculation:

From part (a), the images of the points are $A^{\prime }\left(1,2\right)$ , $B^{\prime }\left(4,6\right)$ , $C^{\prime }\left(6,3\right)$ and $D^{\prime }\left(0,-1\right)$ .

The distance between the points $A\left(0,0\right)$ and $B\left(3,4\right)$ is:

  $\begin{array}{c}AB=\sqrt{{\left(3-0\right)}^2+{\left(4-0\right)}^2}\\ =\sqrt{9+16}\\ =\sqrt{25}\\ =5\end{array}$

The distance between the points $A^{\prime }\left(1,2\right)$ and $B^{\prime }\left(4,6\right)$ is:

  $\begin{array}{c}{A}^{\prime }{B}^{\prime }=\sqrt{{\left(4-1\right)}^2+{\left(6-2\right)}^2}\\ =\sqrt{9+16}\\ =\sqrt{25}\\ =5\end{array}$

The distance between the points $C\left(5,1\right)$ and $D\left(-1,-3\right)$ is:

  $\begin{array}{c}CD=\sqrt{{\left(-1-5\right)}^2+{\left(-3-1\right)}^2}\\ =\sqrt{36+16}\\ =\sqrt{52}\\ =7.2\end{array}$

The distance between the points $C^{\prime }\left(6,3\right)$ and $D^{\prime }\left(0,-1\right)$ is:

  $\begin{array}{c}{C}^{\prime }{D}^{\prime }=\sqrt{{\left(0-6\right)}^2+{\left(-1-3\right)}^2}\\ =\sqrt{36+16}\\ =\sqrt{52}\\ =7.2\end{array}$

Therefore, the length of $AB$ is 5, $A^{\prime }{B}^{\prime }$ is 5, $CD$ is 7.2 and $C^{\prime }{D}^{\prime }$ is 7.2.

c.

To determine

To check: Whether the transformation is an isometry or not.

Answer to Problem 5CE

Yes, the transformation is an isometry.

Explanation of Solution

Given information: The given points are $A\left(0,0\right)$ , $B\left(3,4\right)$ , $C\left(5,1\right)$ and $D\left(-1,-3\right)$ .

From part (b) it can be observed that $AB=A^{\prime }{B}^{\prime }$ and $CD=C^{\prime }{D}^{\prime }$ .

Therefore, the transformation is an isometry.

d.

To determine

To find: The preimage of the points $\left(0,0\right)$ and $\left(4,5\right)$ .

Answer to Problem 5CE

The preimage of the point $\left(0,0\right)$ is $\left(-1,-2\right)$ and $\left(4,5\right)$ is $\left(3,3\right)$ .

Explanation of Solution

Given information: The transformation is $T:\left(x,y\right)\to \left(x+1,y+2\right)$ . The points are $\left(0,0\right)$ and $\left(4,5\right)$ .

Calculation:

The preimage of the point $\left(0,0\right)$ can be calculated as:

  $\begin{array}{c}x+1=0\\ x=-1\end{array}$

And

  $\begin{array}{c}y+2=0\\ y=-2\end{array}$

The coordinate of the preimage of the point $\left(0,0\right)$ is $\left(-1,-2\right)$ .

The preimage of the point $\left(4,5\right)$ can be calculated as:

  $\begin{array}{c}x+1=4\\ x=4-1\\ x=3\end{array}$

And

  $\begin{array}{c}y+2=5\\ y=5-2\\ y=3\end{array}$

The coordinate of the preimage of the point $\left(4,5\right)$ is $\left(3,3\right)$ .

Therefore, the preimage of the point $\left(0,0\right)$ is $\left(-1,-2\right)$ and $\left(4,5\right)$ is $\left(3,3\right)$ .