Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 10 - Section 10.1 - The Rectangular Coordinate System - Exercises - Page 442: 31

Answer

$x_{1}= 0 ;y_{1}=\sqrt{27}+1$ $x_{2}= 0;y_{2}=-\sqrt{27}+1$

Work Step by Step

The two points are on the y axis, so $x_{1}=x_{2}=0$ Using the Distance Formula we can calculate the distances. $6 = \sqrt{(0-3)^{2}+(y-1)^{2}} = \sqrt{9+(y-1)^{2}}$ $9+(y-1)^{2}=36$ $(y-1)^{2}=27$ $(y-1) = \pm\sqrt{27}$ $y=\pm\sqrt{27}+1$ $y_{1}=\sqrt{27}+1$ $y_{2}=-\sqrt{27}+1$
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