Chapter 14.1, Problem 3CE

(a)

To determine

To find: The image of points $P$ and $Q$ .

Answer to Problem 3CE

The images of $P$ and $Q$ are $P^{\prime }\left(2,6\right)$ and $Q^{\prime }\left(8,2\right)$ respectively.

Explanation of Solution

Given information: The mapping is $M:\left(x,y\right)\to \left(2x,2y\right)$ and the diagram is:

  

Calculation:

From the image, the points are $P\left(1,3\right)$ and $Q\left(4,1\right)$ .

To find the images of $P$ , substitute 1 for $x$ and 3 for $y$ in the mapping.

  $\begin{array}{c}M:\left(1,3\right)\to \left(2\cdot 1,2\cdot 3\right)\\ M:\left(1,3\right)\to \left(2,6\right)\\ M:\left(1,3\right)\to P^{\prime }\end{array}$

To find the images of $Q$ , substitute 4 for $x$ and 1 for $y$ in the mapping.

  $\begin{array}{c}M:\left(4,1\right)\to \left(2\cdot 4,2\cdot 1\right)\\ M:\left(4,1\right)\to \left(8,2\right)\\ M:\left(4,1\right)\to Q^{\prime }\end{array}$

Therefore, the images of $P$ and $Q$ are $P^{\prime }\left(2,6\right)$ and $Q^{\prime }\left(8,2\right)$ respectively.

(b)

To determine

To find: Whether the mapping $M$ is a transformation or not.

Answer to Problem 3CE

The mapping $M$ is a transformation.

Explanation of Solution

Given information: The mapping is $M:\left(x,y\right)\to \left(2x,2y\right)$ .

Transformation: A one-to-one mapping from the whole plane to the whole plane.

The given case is also one of the examples of one-to-one mapping because all the pre-image have their images and all the images have their pre-images.

Therefore, the mapping $M$ is a transformation.

(c)

To determine

To find: Whether $M$ appear to be an isometry or not.

Answer to Problem 3CE

The mapping $M$ does not appear to be an isometry.

Explanation of Solution

Given information: The mapping is $M:\left(x,y\right)\to \left(2x,2y\right)$ and the diagram is:

  

Calculation:

From the image, the points are $P\left(1,3\right)$ and $Q\left(4,1\right)$ . Form part (a), the images of $P$ and $Q$ are $P^{\prime }\left(2,6\right)$ and $Q^{\prime }\left(8,2\right)$ respectively.

The formula for the distance between two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by,

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

To find the length of $PQ$ , substitute 4 for $x_2$ , 1 for $x_1$ , 1 for $y_2$ , 3 for $y_1$ in the above formula.

  $\begin{array}{c}PQ=\sqrt{{\left(4-1\right)}^2+{\left(1-3\right)}^2}\\ =\sqrt{{\left(3\right)}^2+{\left(-2\right)}^2}\\ =\sqrt{9+4}\\ =\sqrt{13}\end{array}$

To find the length of $P^{\prime }{Q}^{\prime }$ , substitute 8 for $x_2$ , 2 for $x_1$ , 2 for $y_2$ , 6 for $y_1$ in the above formula.

  $\begin{array}{c}{P}^{\prime }{Q}^{\prime }=\sqrt{{\left(8-2\right)}^2+{\left(2-6\right)}^2}\\ =\sqrt{{\left(6\right)}^2+{\left(-4\right)}^2}\\ =\sqrt{36+16}\\ =\sqrt{52}\end{array}$

The given mapping is not an isometry because $\overline{PQ}\ne \overline{P^{\prime }{Q}^{\prime }}$ .

Therefore, the mapping $M$ does not appear to be an isometry.

(d)

To determine

To decide: Whether $M$ maps the mid-point of $\overline{PQ}$ to the midpoint of $\overline{P^{\prime }{Q}^{\prime }}$ .

Answer to Problem 3CE

The mapping $M$ does map to the midpoint $\overline{PQ}$ to the midpoint of $\overline{P^{\prime }{Q}^{\prime }}$ .

Explanation of Solution

Given information: The mapping is $M:\left(x,y\right)\to \left(2x,2y\right)$ .

Calculation:

Let $N$ is the midpoint of the line $PQ$ and $N^{\prime }$ is the midpoint of the line $P^{\prime }{Q}^{\prime }$ . The points are $P\left(1,3\right)$ and $Q\left(4,1\right)$ . The images of points are $P^{\prime }\left(2,6\right)$ and $Q^{\prime }\left(8,2\right)$ respectively.

The point $N$ can be calculated as:

  $\begin{array}{c}N=\left(\frac{1+4}{2},\frac{3+1}{2}\right)\\ =\left(\frac{5}{2},2\right)\end{array}$

The point $N^{\prime }$ can be calculated as:

  $\begin{array}{c}{N}^{\prime }=\left(\frac{2+8}{2},\frac{6+2}{2}\right)\\ =\left(5,4\right)\end{array}$

Now, from the mapping image of $N$ ,

  $\begin{array}{c}M:\left(\frac{5}{2},2\right)\to \left(2\cdot \frac{5}{2},2\cdot 2\right)\\ M:\left(\frac{5}{2},2\right)\to \left(5,4\right)\\ M:\left(\frac{5}{2},2\right)\to N^{\prime }\end{array}$

Therefore, the mapping $M$ does map to the midpoint $\overline{PQ}$ to the midpoint of $\overline{P^{\prime }{Q}^{\prime }}$ .