Question

Use the transformation $T:\left(x,y\right)\to \left(x+1,y+2\right)$ in this exercise.

1. Plot the following points and their images on the chalkboard: $A\left(0,0\right)$, $B\left(3,4\right)$, $C\left(5,1\right)$, and $D\left(-1,-3\right)$.
2. Find $AB$, $A\text{'}B\text{'}$, $CD$, and, $C\text{'}D\text{'}$.
3. Does this transformation appear to be an isometry?
4. What is the pre image of $\left(0,0\right)$? of $\left(4,5\right)$?

Short answer

(a) The points and images of their transformation is shown below.

(b) The length of $AB$ is 5, $A\text{'}B\text{'}$ is 5, $CD$ is 7.2 and $C\text{'}D\text{'}$ is 7.2.

(c) Yes, the transformation is an isometry.

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

Step-by-step solution

a.Step 1. 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 $D\left(-1,-3\right)$.

Step 2. Calculation.

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.

Step 3. Conclusion.

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

b.Step 1. Given Information.

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

Step 2. Formula Used.

The formula to find the distance between two points $\left(x_1,x_2\right)$ and $\left(y_1,y_2\right)$ is:

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

Step 3. Explanation.

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

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 $A\text{'}\left(1,2\right)$ and $B\text{'}\left(4,6\right)$ is:

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

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 $C\text{'}\left(6,3\right)$ and $D\text{'}\left(0,-1\right)$ is:

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

Step 4. Conclusion.

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

c.Step 1. Given Information.

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

Step 2. Calculation.

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

Step 3. Conclusion.

Therefore, the transformation is an isometry.

d.Step 1. 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 $\left(4,5\right)$.

Step 2. 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 $\left(3,3\right)$.

Step 3. Conclusion.

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