## Calculus 10th Edition

Published by Brooks Cole

# Chapter P - P.2 - Linear Models and Rates of Change - Exercises - Page 17: 61

#### Answer

a.$24y-40x+9=0$ b. $40y+24x-53=0$

#### Work Step by Step

Put the equation into slope-intercept form. $5x-3y=0$ $3y=5x$ $y=\frac{5}{3}x$ The slope is $\frac{5}{3}$. a. Parallel lines have the same slope. $y=mx+b$ $y=\frac{5}{3}x+b$ Plug in $(\frac{3}{4}, \frac{7}{8})$ to find b. $\frac{7}{8}=\frac{5}{3}(\frac{3}{4})+b$ $\frac{7}{8}=\frac{5}{4}+b$ $\frac{7}{8}=\frac{10}{8}+b$ $b=-\frac{3}{8}$ $y=\frac{5}{3}x-\frac{3}{8}$ $y-\frac{5}{3}x+\frac{3}{8}$ Multiply both sides by 24. $24y-40x+9=0$ b. The slope of a perpendicular line is the negative reciprocal of the original slope, so the slope for this line is $-\frac{3}{5}$. $y=-\frac{3}{5}x+b$ Plug in $(\frac{3}{4}, \frac{7}{8})$ to find b. $\frac{7}{8}=-\frac{3}{5}*\frac{3}{4}+b$ $\frac{7}{8}=-\frac{9}{20}+b$ $\frac{70}{80}=-\frac{36}{80}+b$ $\frac{106}{80}=b$ $b=\frac{53}{40}$ $y=-\frac{3}{5}x+\frac{53}{40}$ $y+\frac{3}{5}x-\frac{53}{40}$ Multiply both sides by 40. $40y+24x-53=0$

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