Chapter 14.1, Problem 8WE

(a).

To determine

To graph: The three points $A\left(0,4\right)$ , $B\left(4,6\right)$ and $C\left(2,0\right)$ and their images under the given transformation.

Explanation of Solution

Given information: The given transformation is $H:\left(x,y\right)\to \left(-x,-y\right)$ .

Graph:

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

  $H:\left(x,y\right)\to \left(-x,-y\right)$

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

  $\left(-\left(0\right),-\left(4\right)\right)=\left(0,-4\right)$

Therefore, image of point $A\left(0,4\right)$ is $A^{\prime }\left(0,-4\right)$ .

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

  $\left(-\left(4\right),-\left(6\right)\right)=\left(-4,-6\right)$

Therefore, image of point $B\left(4,6\right)$ is $B^{\prime }\left(-4,-6\right)$ .

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

  $\left(-\left(2\right),-\left(0\right)\right)=\left(-2,0\right)$

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

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

  

Interpretation:

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

(b).

To determine

To find: Whether the given transformation appears to be an isometric or not.

Answer to Problem 8WE

The given transformation appears to be an isometric.

Explanation of Solution

Given information: The given transformation is $H:\left(x,y\right)\to \left(-x,-y\right)$ .

Calculation:

As calculated in the above graph, $A^{\prime }\left(0,-4\right)$ is the image of point $A\left(0,4\right)$ , $B^{\prime }\left(-4,-6\right)$ is the image of point $B\left(4,6\right)$ and $C^{\prime }\left(-2,0\right)$ 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 $\overline{AB}\cong \overline{A^{\prime }{B}^{\prime }}$

, $\overline{BC}\cong \overline{B^{\prime }{C}^{\prime }}$ and $\overline{AC}\cong \overline{A^{\prime }{C}^{\prime }}$ .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 $A^{\prime }\left(0,-4\right)=\left(x_1,y_1\right)$ and $B^{\prime }\left(-4,-6\right)=\left(x_2,y_2\right)$ .

To determine the distance $A^{\prime }{B}^{\prime }$ , substitute $0$ for $x_1$ , $-4$ for $y_1$ , $-4$ for $x_2$ and $-6$ for $y_2$ in the expression for distance formula.

  $\begin{array}{c}{A}^{\prime }{B}^{\prime }=\sqrt{{\left(-4-0\right)}^2+{\left(-6-\left(-4\right)\right)}^2}\\ =\sqrt{{\left(-4\right)}^2+{\left(-2\right)}^2}\\ =\sqrt{16+4}\\ =\sqrt{20}\end{array}$

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 $B^{\prime }\left(-4,-6\right)=\left(x_1,y_1\right)$ and $C^{\prime }\left(-2,0\right)=\left(x_2,y_2\right)$ .

To determine the distance $B^{\prime }{C}^{\prime }$ , substitute $-4$ for $x_1$ , $-6$ for $y_1$ , $-2$ for $x_2$ and $0$ for $y_2$ in the expression for distance formula.

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

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 $A^{\prime }\left(0,-4\right)=\left(x_1,y_1\right)$ and $C^{\prime }\left(-2,0\right)=\left(x_2,y_2\right)$ .

To determine the distance $A^{\prime }{C}^{\prime }$ , substitute $0$ for $x_1$ , $-4$ for $y_1$ , $-2$ for $x_2$ and $0$ for $y_2$ in the expression for distance formula.

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

Therefore, from the above it is prove that $\overline{AB}=\overline{A^{\prime }{B}^{\prime }}$ , $\overline{BC}=\overline{B^{\prime }{C}^{\prime }}$ and $\overline{AC}=\overline{A^{\prime }{C}^{\prime }}$ . Since, they have the equal distance, this implies that two triangle are congruent and thus isometric.

(c).

To determine

To find: The coordinates of the preimage of the given point.

Answer to Problem 8WE

The coordinates of the preimage of the given point is $\left(4,2\right)$ .

Explanation of Solution

Given information: The given transformation is $H:\left(x,y\right)\to \left(-x,-y\right)$ and the point of which preimage is to be determined is $\left(12,6\right)$ .

Calculation:

The given transformation is $H:\left(x,y\right)\to \left(-x,-y\right)$ and $H\left(x,y\right)\to \left(12,6\right)$ .

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

  $\begin{array}{c}-x=12\\ x=-12\end{array}$

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

  $\begin{array}{c}-y=6\\ y=-6\end{array}$

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