Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 2 - Section 2.5 - The Point-Slope Form of the Equation of a Line - Exercise Set - Page 164: 60

Answer

$f(x)=-\frac{1}{3}x+\frac{13}{3}$

Work Step by Step

Slope of the perpendicular line:- $x-$intercept is $3$. The point on line is $(3,0)$. $y-$intercept is $-9$. The point on line is $(0,-9)$. Slope of the line passes through $(3,0)$ and $(0,-9)$ is $\Rightarrow m_1=\frac{Change \; in \; y}{Change \; in \; x}$ $\Rightarrow m_1=\frac{-9-0}{0-3}$ Simplify. $\Rightarrow m_1=\frac{-9}{-3}$ $\Rightarrow m_1=3$. Two lines are perpendicular if their slopes are negative reciprocal to each other. Hence, slope of the perpendicular line is $m_2=−\frac{1}{m_1}$ $m_2=−\frac{1}{3}$ The standard equation of a line in a slope intercept form is $\Rightarrow y=mx+c$ Where, $m=$ slope and $c=$ $y-$ intercept. Substitute the value of slope. $\Rightarrow y=(-\frac{1}{3})x+c$ We are given that $f$ passes through $(-5,6)$. Plug $x=-5$ and $y=6$ into the line equation. $\Rightarrow 6=(-\frac{1}{3})(-5)+c$ Simplify. $\Rightarrow 6=\frac{5}{3}+c$ Subtract $\frac{5}{3}$ from both sides. $\Rightarrow 6-\frac{5}{3}=\frac{5}{3}+c-\frac{5}{3}$ Simplify. $\Rightarrow \frac{13}{3}=c$ Substitute the value of $c$ into the line equation. $\Rightarrow y=(-\frac{1}{3})x+\frac{13}{3}$ Plug $y=f(x)$. $\Rightarrow f(x)=-\frac{1}{3}x+\frac{13}{3}$
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