Algebra: A Combined Approach (4th Edition)

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

Chapter 8 - Section 8.1 - Review of Equations of Lines and Writing Parallel and Perpendicular Lines - Exercise Set: 15

Answer

y+8=$\frac{1}{3}$(x+3) 7=$\frac{1}{3}$x-y

Work Step by Step

(-3;-8),(-6;-9) m=$\frac{-9+8}{-6+3}$=$\frac{-1}{-3}$=$\frac{1}{3}$ slope $\frac{1}{3}$ ,through(-3;-8) y-$y_{1}$=m(x-$x_{1}$) y+8=$\frac{1}{3}$(x+3) y+8=$\frac{1}{3}$x+1 8-1=$\frac{1}{3}$x-y 7=$\frac{1}{3}$x-y
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