91Ó°ÊÓ

The points are $A\left(6,1\right)$, $B\left(3,4\right)$, and $C\left(1,-3\right)$, and the transformation is $R:\left(x,y\right)→\left(-x,y\right)$.

The images $A\text{'}$, $B\text{'}$, and $C\text{'}$ under the transformation $R:\left(x,y\right)→\left(-x,y\right)$ are as follows:

$\begin{array}{l}T:A\left(6,\text{ }1\right)→A^{\text{'}}\left(−6,\text{ }1\right)\\ T:B\left(3,\text{ }4\right)→B^{\text{'}}\left(−3,\text{ }4\right)\\ T:C\left(1,\text{ }−3\right)→C^{\text{'}}\left(−1,\text{ }−3\right)\end{array}$

Now, plot all the points and their images on the coordinate plane as shown below.

From the graph it can be observed that the points A, B, C and their images $A\text{'}$, $B\text{'}$, $C\text{'}$ are at same distance from the y-axis.

The points are $A\left(6,1\right)$, $B\left(3,4\right)$, and $C\left(1,-3\right)$. The transformation is $R:\left(x,y\right)→\left(-x,y\right)$.

Let $P\left(x_1,y_1\right)$ and $Q\left(x_2,y_2\right)$ be any two points. The image of the points under the transformation $R:\left(x,y\right)→\left(-x,y\right)$ is:

$\begin{array}{l}R:P\left(x_1,\text{ }{y}_1\right)→P^{\text{'}}\left(−x_1,\text{ }{y}_1\right)\\ R:Q\left(x_2,\text{ }{y}_2\right)→Q^{\text{'}}\left(−x_2,\text{ }{y}_2\right)\end{array}$

To find the distance between two points, use the following distance formula.

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

The distance between points P and Q is:

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

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

$\begin{array}{c}{P}^{\text{'}}{Q}^{\text{'}}=\sqrt{{\left(\left(−x_2\right)−\left(−x_1\right)\right)}^2+{\left(y_2−y_1\right)}^2}\\ =\sqrt{{\left(−\left[x_2−x_1\right]\right)}^2+{\left(y_2−y_1\right)}^2}\\ =\sqrt{{\left(x_2−x_1\right)}^2+{\left(y_2−y_1\right)}^2}\end{array}$

So, $PQ=P\text{'}Q\text{'}$.

Hence, it is proved that the transformation R is an isometry.