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Q23.

Expert-verifiedFound in: Page 576

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

- it is proved that the transformation
is*R***an****isometry**.

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

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:

$\begin{array}{l}T:A\left(6,\text{\hspace{0.33em}}1\right)\to {A}^{\text{'}}\left(-6,\text{\hspace{0.33em}}1\right)\\ T:B\left(3,\text{\hspace{0.33em}}4\right)\to {B}^{\text{'}}\left(-3,\text{\hspace{0.33em}}4\right)\\ T:C\left(1,\text{\hspace{0.33em}}-3\right)\to {C}^{\text{'}}\left(-1,\text{\hspace{0.33em}}-3\right)\end{array}$

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

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

$\begin{array}{l}R:P\left({x}_{1},\text{\hspace{0.33em}}{y}_{1}\right)\to {P}^{\text{'}}\left(-{x}_{1},\text{\hspace{0.33em}}{y}_{1}\right)\\ R:Q\left({x}_{2},\text{\hspace{0.33em}}{y}_{2}\right)\to {Q}^{\text{'}}\left(-{x}_{2},\text{\hspace{0.33em}}{y}_{2}\right)\end{array}$

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.

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