Question

For each transformation given:

1. Plot the three points $A\left(0,4\right)$, $B\left(4,6\right)$, and $C\left(2,0\right)$, and their images $A\text{'}$, $B\text{'}$, and $C\text{'}$ under the transformation.
2. State whether the transformation appears to be an isometry.
3. Find the pre image of $\left(12,6\right)$.

$G:\left(x,y\right)\to \left(-\frac{1}{2}x,-\frac{1}{2}y\right)$

Short answer

1. The given transformation appears not to be an isometric.
2. The coordinates of the preimage of the given point is $\left(-24,-12\right)$.

Step-by-step solution

a.Step 1. Given Information.

The given transformation is $G:\left(x,y\right)\to \left(-\frac{1}{2}x,-\frac{1}{2}y\right)$.

Step 2. Explanation.

To find the images of the given points, consider the given transformation:

$G:\left(x,y\right)\to \left(-\frac{1}{2}x,-\frac{1}{2}y\right)$

To determine image of point $A\left(0,4\right)$, substitute 0 for x and 4 for y in the above expression.

$\left(-\frac{1}{2}\left(0\right),-\frac{1}{2}\left(4\right)\right)=\left(0,-2\right)$

So, image of point $A\left(0,4\right)$ is $A\text{'}\left(0,-2\right)$.

To determine image of point $B\left(4,6\right)$, substitute 4 for x and 6 for y in the above expression.

$\left(-\frac{1}{2}\left(4\right),-\frac{1}{2}\left(6\right)\right)=\left(-2,-3\right)$

Therefore, image of point $B\left(4,6\right)$ is $B\text{'}\left(-2,-3\right)$.

To determine image of point $C\left(2,0\right)$, substitute 2 for x and 0 for y in the above expression.

$\left(-\frac{1}{2}\left(2\right),-\frac{1}{2}\left(0\right)\right)=\left(-1,0\right)$

So, the image of point $C\left(2,0\right)$ is $C\text{'}\left(-1,0\right)$.

Plot and connect the ordered pair $C\text{'}\left(-1,0\right)$, $B\left(4,6\right)$ and $C\left(2,0\right)$, and their images $A\text{'}\left(0,-2\right)$, $B\text{'}\left(-2,-3\right)$ and $C\text{'}\left(-1,0\right)$ on the coordinate graph as shown below:

Step 3. Conclusion.

In the above graph, $A\text{'}\left(0,-2\right)$is the image of point $A\left(0,4\right)$, $B\text{'}\left(-2,-3\right)$ is the image of point $B\left(4,6\right)$ and $C\text{'}\left(-1,0\right)$ is the image of point $C\left(2,0\right)$.

b.Step 1. Given Information.

The given transformation is $G:\left(x,y\right)\to \left(-\frac{1}{2}x,-\frac{1}{2}y\right)$.

Step 2. Calculation.

As calculated in the above graph, $A\text{'}\left(0,-2\right)$is the image of point $A\left(0,4\right)$, $B\text{'}\left(-2,-3\right)$is the image of point $B\left(4,6\right)$and $C\text{'}\left(-1,0\right)$is the image of point $C\left(2,0\right)$.

In the theorem of transformations, an isometry maps a triangle to a congruent triangle.

For mapping the distances between image and point must be same such as $\overline{AB}\cong \overline{A\text{'}B\text{'}}$, $\overline{BC}\cong \overline{B\text{'}C\text{'}}$and $\overline{AC}\cong \overline{A\text{'}C\text{'}}$.If they have the equal distance, then this implies that two triangle are congruent and thus isometric.

The expression for distance formula is:

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

Let, the point of $A\left(0,4\right)=\left(x_1,y_1\right)$and $B\left(4,6\right)=\left(x_2,y_2\right)$.

To determine the distance AB, substitute 0 for $x_1$, 4 for $y_1$, 4 for $x_2$and 6 for$y_2$ in the expression for distance formula.
$\begin{array}{c}AB=\sqrt{{\left(4-0\right)}^2+{\left(6-4\right)}^2}\\ =\sqrt{{\left(4\right)}^2+{\left(2\right)}^2}\\ =\sqrt{16+4}\\ =\sqrt{20}\end{array}$

Let, the point of $A\text{'}\left(0,-2\right)=\left(x_1,y_1\right)$and $B\text{'}\left(-2,-3\right)=\left(x_2,y_2\right)$.

To determine the distance $B\text{'}\left(-2,-3\right)=\left(x_2,y_2\right)$, substitute 0 for $x_1$, -2 for $y_1$, -2 for $x_2$and -3 for $y_2$in the expression for distance formula.

$\begin{array}{c}{A}^{\text{'}}{B}^{\text{'}}=\sqrt{{\left(-2-0\right)}^2+{\left(-3-\left(-2\right)\right)}^2}\\ =\sqrt{{\left(-2\right)}^2+{\left(-1\right)}^2}\\ =\sqrt{4+1}\\ =\sqrt{5}\end{array}$

Let, the point of $B\left(4,6\right)=\left(x_1,y_1\right)$and $C\left(2,0\right)=\left(x_2,y_2\right)$.

To determine the distance $BC$, substitute 4 for $x_1$, 6 for $y_1$, 2 for $x_2$and 0 for $y_2$in the expression for distance formula.

$\begin{array}{c}BC=\sqrt{{\left(2-4\right)}^2+{\left(0-6\right)}^2}\\ =\sqrt{{\left(-2\right)}^2+{\left(-6\right)}^2}\\ =\sqrt{4+36}\\ =\sqrt{40}\end{array}$

Let, the point of $B\text{'}\left(-2,-3\right)=\left(x_1,y_1\right)$and $C\text{'}\left(-1,0\right)=\left(x_2,y_2\right)$.

To determine the distance $B\text{'}C\text{'}$, substitute -2 for $x_1$, -3 for $y_1$, -1 for $x_2$ and 0 for $y_2$ in the expression for distance formula.

$\begin{array}{c}{B}^{\text{'}}{C}^{\text{'}}=\sqrt{{\left(-1-\left(-2\right)\right)}^2+{\left(0-\left(-3\right)\right)}^2}\\ =\sqrt{{\left(1\right)}^2+{\left(3\right)}^2}\\ =\sqrt{1+9}\\ =\sqrt{10}\end{array}$

Let, the point of $A\left(0,4\right)=\left(x_1,y_1\right)$ and $C\left(2,0\right)=\left(x_2,y_2\right)$.

To determine the distance $AC$, substitute 0 for $x_1$, 4 for $y_1$, 2 for $x_2$ and 0 for $y_2$ in the expression for distance formula.

$\begin{array}{c}AC=\sqrt{{\left(2-0\right)}^2+{\left(0-4\right)}^2}\\ =\sqrt{{\left(2\right)}^2+{\left(-4\right)}^2}\\ =\sqrt{4+16}\\ =\sqrt{20}\end{array}$

Let, the point of $A\text{'}\left(0,-2\right)=\left(x_1,y_1\right)$ and $C\text{'}\left(-1,0\right)=\left(x_2,y_2\right)$.

To determine the distance $A\text{'}C\text{'}$, substitute 0 for $x_1$, -2 for $y_1$, -1 for $x_2$ and 0 for $y_2$ in the expression for distance formula.

$\begin{array}{c}{A}^{\text{'}}{C}^{\text{'}}=\sqrt{{\left(-1-0\right)}^2+{\left(0-\left(-2\right)\right)}^2}\\ =\sqrt{{\left(-1\right)}^2+{\left(2\right)}^2}\\ =\sqrt{1+4}\\ =\sqrt{5}\end{array}$

Step 3. Conclusion.

Therefore, from the above it is prove that $\overline{AB}\ne \overline{A\text{'}B\text{'}}$, $\overline{BC}\ne \overline{B\text{'}C\text{'}}$ and $\overline{AC}\ne \overline{A\text{'}C\text{'}}$. Since, they don’t have the equal distance, this implies that two triangle are not congruent and thus not isometric.

c.Step 1. Given Information.

The given transformation is $G:\left(x,y\right)\to \left(-\frac{1}{2}x,-\frac{1}{2}y\right)$ and the point of which preimage is to be determined is $\left(12,6\right)$.

Step 2. Calculation.

The given transformation is $G:\left(x,y\right)\to \left(-\frac{1}{2}x,-\frac{1}{2}y\right)$ and $G\left(x,y\right)\to \left(12,6\right)$

On comparing the two values of $x-\text{axis}$:

$\begin{array}{c}-\frac{1}{2}x=12\\ x=12\times \left(-2\right)\\ =-24\end{array}$

On comparing the two values of $y-\text{axis}$:

$\begin{array}{c}-\frac{1}{2}y=6\\ y=6\times \left(-2\right)\\ =-12\end{array}$

Step 3. Conclusion.

Therefore, the coordinates of the preimage of $\left(12,6\right)$ is $\left(-24,-12\right)$.