91Ó°ÊÓ

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

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.

$\begin{array}{c}M:\left(1,3\right)→\left(2â‹\dots 1,2â‹\dots 3\right)\\ M:\left(1,3\right)→\left(2,6\right)\\ M:\left(1,3\right)→P^{\text{'}}\end{array}$

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

$\begin{array}{c}M:\left(4,1\right)→\left(2â‹\dots 4,2â‹\dots 1\right)\\ M:\left(4,1\right)→\left(8,2\right)\\ M:\left(4,1\right)→Q^{\text{'}}\end{array}$

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

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

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.

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

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 $P\text{'}\left(2,6\right)$ and $Q\text{'}\left(8,2\right)$ 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 $P\text{'}Q\text{'}$, substitute 8 for $x_2$, 2 for $x_1$, 2 for $y_2$, 6 for $y_1$ in the above formula.

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

The given mapping is not an isometry because $\stackrel{¯}{PQ}≠\stackrel{¯}{P\text{'}Q\text{'}}$.

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

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

Let N is the midpoint of the line PQ and $N\text{'}$is the midpoint of the line$P\text{'}Q\text{'}$. The points are $P\left(1,3\right)$and $Q\left(4,1\right)$. The images of points are $P\text{'}\left(2,6\right)$and $Q\text{'}\left(8,2\right)$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 $N\text{'}$can be calculated as:

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

Now, from the mapping image of N,

$\begin{array}{c}M:\left(\frac{5}{2},2\right)→\left(2â‹\dots \frac{5}{2},2â‹\dots 2\right)\\ M:\left(\frac{5}{2},2\right)→\left(5,4\right)\\ M:\left(\frac{5}{2},2\right)→N^{\text{'}}\end{array}$

Therefore, the mapping M does map to the midpoint $\stackrel{¯}{PQ}$ to the midpoint of $\stackrel{¯}{P\text{'}Q\text{'}}$.