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Q23.

Expert-verified

Geometry

Found in: Page 576

Book edition Student Edition

Author(s) Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

Pages 227 pages

ISBN 9780395977279

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Short Answer

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

it is proved that the transformation R is an isometry.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

a.Step 1. Given Information.

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

Step 2. Explanation.

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

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

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 $A'$ , $B'$ , $C'$ are at same distance from the y-axis.

b.Step 1. Given Information.

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

Step 2. Proof.

Let $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ be any two points. The image of the points under the transformation $R : (x, y) \to (- x, y)$ is:

$\begin{array}{l} R : P (x_{1}, y_{1}) \to P^{'} (- x_{1}, y_{1}) \\ R : Q (x_{2}, y_{2}) \to Q^{'} (- x_{2}, y_{2}) \end{array}$

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

$D = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

The distance between points P and Q is:

$P Q = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

The distance between points $P'$ and $Q'$ is:

$\begin{matrix} P^{'} Q^{'} = \sqrt{{((- x_{2}) - (- x_{1}))}^{2} + {(y_{2} - y_{1})}^{2}} \\ = \sqrt{{(- [x_{2} - x_{1}])}^{2} + {(y_{2} - y_{1})}^{2}} \\ = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} \end{matrix}$

So, $P Q = P' Q'$ .

Step 3. Conclusion.

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

Most popular questions for Math Textbooks

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Is S a transformation? Explain.

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