Elementary Geometry for College Students (5th Edition)

Published by Brooks Cole
ISBN 10: 1439047901
ISBN 13: 978-1-43904-790-3

Chapter 10 - Section 10.2 - Graphs of Linear Equations and Slope - Exercises - Page 465: 34

Answer

$x=6,-1$

Work Step by Step

To be a right triangle, one set of slopes must be opposite reciprocals, for this would make two of the lines perpendicular. Thus: $m_1 = \frac{y_2-y_1}{x_2-x_1} = \frac{x-2}{4-2} = (x-2)/4$ $m_2 = \frac{y_2-y_1}{x_2-x_1} = \frac{x-3}{4-7} =(x-3)/(-3) $ We set the opposite reciprocals equal to each other: $ \frac{-4}{x-2} = \frac{x-3}{-3} \\ 12 = x^2 -5x +6 \\ 0 = x^2 -5x -6 \\ 0= (x-6)(x+1)\\ x= 6, -1$
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