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Q23.

Expert-verified
Found in: Page 576

### Geometry

Book edition Student Edition
Author(s) Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen
Pages 227 pages
ISBN 9780395977279

# Plot the points $A\left(6,1\right),B\left(3,4\right)$, and $C\left(1,-3\right)$ and their images $A\text{'},B\text{'}$, and $C\text{'}$ under the transformation $R:\left(x,y\right)\to \left(-x,y\right)$.Prove that R is an isometry. (Hint: Let $P\left({x}_{1},{y}_{1}\right)$ and $Q\left({x}_{2},{y}_{2}\right)$ be any two points. Find $P\text{'}$ and $Q\text{'}$, and use the distance formula to show that $PQ=P\text{'}Q\text{'}$.)

1. it is proved that the transformation R is an isometry.
See the step by step solution

## a.Step 1. Given Information.

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

## Step 2. Explanation.

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

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

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

## Step 3. Conclusion.

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.

## b.Step 1. Given Information.

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

## Step 2. Proof.

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

$\begin{array}{l}R:P\left({x}_{1},\text{ }{y}_{1}\right)\to {P}^{\text{'}}\left(-{x}_{1},\text{ }{y}_{1}\right)\\ R:Q\left({x}_{2},\text{ }{y}_{2}\right)\to {Q}^{\text{'}}\left(-{x}_{2},\text{ }{y}_{2}\right)\end{array}$

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

## Step 3. Conclusion.

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