Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 3 - Sections 3.1-3.5 - Integrated Review - Linear Equations in Two Variables - Page 176: 21



Work Step by Step

Using $y=mx+b$ where $m$ is the slope, the given equation, $ 4x+y=\dfrac{2}{3} ,$ is equivalent to \begin{array}{l}\require{cancel} y=-4x+\dfrac{2}{3} .\end{array} Hence, the slope is $m=-4$. Since perpendicular lines have negative reciprocal slopes, then $m_p=\dfrac{1}{4}$ Using $ (2,-3) $ and $m_p= \dfrac{1}{4} ,$ the equation of the line is \begin{array}{l}\require{cancel} y-(-3)=\dfrac{1}{4}(x-2) \\\\ y+3=\dfrac{1}{4}(x-2) \\\\ y+3=\dfrac{1}{4}x-\dfrac{1}{2} \\\\ y=\dfrac{1}{4}x-\dfrac{1}{2}-3 \\\\ y=\dfrac{1}{4}x-\dfrac{1}{2}-\dfrac{6}{2} \\\\ y=\dfrac{1}{4}x-\dfrac{7}{2} .\end{array} In function notation form, this is equivalent to $ f(x)=\dfrac{1}{4}x-\dfrac{7}{2} .$
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