Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 10 - Section 10.1 - The Rectangular Coordinate System - Exercises - Page 450: 32

Answer

$x_{1}=\sqrt{35}+3; y_{1}=0$ $x_{2}=-\sqrt{35}+3; y_{2}=0$

Work Step by Step

The two points are on the x axis, so $y_{1}=y_{2}=0$ Using the Distance Formula we can calculate the distances. $\sqrt{(x-3)^{2}+(0-1)^{2}} = \sqrt{(x-3)^{2}+1} = 6$ $(x-3)^{2}+1=36$ $(x-3)^{2}=35$ $(x-3) = \pm\sqrt{35}$ $x=\pm\sqrt{35}+3$ $x_{1}=\sqrt{35}+3$ $x_{2}=-\sqrt{35}+3$
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