Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

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


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).
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.