Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 3 - Section 3.4 - Slope and Rate of Change - Exercise Set: 48

Answer

$m_{parallel}=\frac{9}{10}$ $m_{perpendicular}=-\frac{10}{9}$

Work Step by Step

A line parallel to the line that passes through the points (6, -1) and (-4, -10) will have a slope of $m_{parallel}=\frac{-10-(-1)}{-4-(6)}=\frac{-9}{-10}=\frac{9}{10}$ A line perpendicular to the line that passes through the points (-8, -4) and (3, 5) will have a slope of $m_{perpendicular}=-\frac{10}{9}$
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