91Ó°ÊÓ

The given point are $P\left(4,5\right)$, $Q\left(-3,2\right)$and $R\left(-3,-1\right)$. The mapping is $S:\left(x,y\right)→\left(x,0\right)$.

Consider the mapping $S:\left(x,y\right)→\left(x,0\right)$.

The image of point $P\left(4,5\right)$is:

$P:\left(4,5\right)→P\text{'}\left(4,0\right)$

The image of point $Q\left(-3,2\right)$is:

$Q:\left(-3,2\right)→Q\text{'}\left(-3,0\right)$

The image of point $R\left(-3,-1\right)$is:

$R:\left(-3,-1\right)→R\text{'}\left(-3,0\right)$

Figure (1)

From the graph it can be observed that the distance between the point and its image is the x-coordinate of the point.

The given point are $P\left(4,5\right)$, $Q\left(-3,2\right)$ and $R\left(-3,-1\right)$. The mapping is $S:\left(x,y\right)→\left(x,0\right)$.

From part (a), the images of the given points are $P\text{'}\left(4,0\right)$, $Q\text{'}\left(-3,0\right)$ and $R\text{'}\left(-3,0\right)$.

For isometry the distance between $P,Q,R$ and $P\text{'},Q\text{'},R\text{'}$ must be the same.

The distance between the point P and Q is:

$\begin{array}{c}PQ=\sqrt{{\left(4+3\right)}^2+{\left(5−2\right)}^2}\\ =\sqrt{7^2+3^2}\\ =\sqrt{49+9}\\ =\sqrt{58}\end{array}$

The distance between the point $P\text{'}$ and $Q\text{'}$ is:

$\begin{array}{c}P\text{'}Q\text{'}=\sqrt{{\left(4+3\right)}^2+{\left(0−0\right)}^2}\\ =\sqrt{7^2+0}\\ =\sqrt{49}\\ =7\end{array}$

Since $PQ≠P\text{'}Q\text{'}$. So, mapping $S$ is not an isometry.

Therefore, the mapping S does not appear an isometry.

The given mapping is $S:\left(x,y\right)→\left(x,0\right)$.

If the image of origin is origin, then the mapping is said to be transformation. The image of origin is:

$S:\left(0,0\right)→\left(0,0\right)$

Hence, the mapping S is a transformation.