## Calculus 10th Edition

Published by Brooks Cole

# Chapter P - P.2 - Linear Models and Rates of Change - Exercises - Page 17: 60

#### Answer

a.) $\frac{7}{4}x+y-\frac{23}{24}=0$ b.)$-\frac{4}{7}x+y+\frac{41}{42}=0$

#### Work Step by Step

The slope needs to be found first, which means solving the given equation for the variable 'y'. $7x + 4y = 8$ $4y = -7x + 8$ Subtract 7x $y = -\frac{7}{4}x + \frac{8}{4}$ Divide by 4, isolating y This is in the form y=mx+b, so m, the slope, is equal to -7/4. A line parallel to the line given will have slope -7/4, and will pass through the point $(\frac{5}{6},\frac{-1}{2})$, as given by the problem. There is enough information to plug this into the point slope equation, $y-y_1 =m(x-x_1)$. THis will then be put into the general form $Ax + By + C = 0$ a.) Parallel line equation: $y-(-\frac{1}{2}) = -\frac{7}{4}(x-\frac{5}{6})$ Distribute 'm', and put it into the general form: $\frac{7}{4}x+y-\frac{23}{24}=0$ This is the answer to (a). (b) is solved in a similar manner, except the slope is different since the new line is perpendicular to the original. A perpendicular line has a slope of the negative reciprocal of the original slope. $m_0=-\frac{7}{4}$ $m_{perp} = \frac{4}{7}$ Now plug this value into the same point slope equation, and solve for the general form. Point slope form: $y-(-\frac{1}{2}) = \frac{4}{7}(x-\frac{5}{6})$ Put in general form: $-\frac{4}{7}x+y+\frac{41}{42}=0$ This is the answer to (b).

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