Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 2 - Graphs and Functions - 2.5 Equations of Lines and Linear Models - 2.5 Exercises - Page 244: 51


(a) Standard Form: $x+3y=11$ (b) Slope-intercept form: $y=-\frac{1}{3}x+\frac{11}{3}$

Work Step by Step

RECALL: (1) The standard form of a line's equation is $Ax+By=C$. (2) THe slope-intercept form of a line's equation is $y=mx+b$ where $m$=slope and $(0, b)$ is the line's y-intercept. (3) Parallel lines have the same slope. (4) Perpendicular lines have slopes whose product is $-1$ (negative reciprocals of each other). The line is parallel to $x+3y=5$. Solve the equation for $y$ to obtain: $3y=-x+5 \\\frac{3y}{3}=\frac{-x+5}{3} \\y=-\frac{1}{3}x+\frac{5}{3}$ The slope of the line above is $-\frac{1}{3}$. Thus, the parallel of the line parallel to it is also $-\frac{1}{3}$. This means that the tentative equation of the line we are looking for is $y=-\frac{1}{3}x+b$. To find the value of $b$, substitute the $x$ and $y$ values of the point $(-1, 4)$ to obtain: $y=-\frac{1}{3}x+b \\4=-\frac{1}{3}(-1)+b \\4=\frac{1}{3}+b \\4-\frac{1}{3}=b \\\frac{12}{3}-\frac{1}{3} =b \\\frac{11}{3}=b$ Therefore, the equation of the line is $y=-\frac{1}{3}x+\frac{11}{3}$. (a) Standard form Multiply $3$ to both sides of the equation to obtain: $3(y)=3(-\frac{1}{3}x+\frac{11}{3}) \\3y=-x+11$ Add $x$ to both sides of the equation to obtain: $x+3y=-x+11+x \\x+3y=11$ (b) Slope-intercept form $y=-\frac{1}{3}x+\frac{11}{3}$
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