Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.10 - Lines - 1.10 Exercises - Page 115: 40

Answer

$y=2x+8$, or, in general form, $2x-y+8=0$

Work Step by Step

Through $(-3,2);$ perpendicular to the line $y=-\frac{1}{2}x+7$ Use the point-slope form of the equation of a line, which is $y-y_{1}=m(x-x_{1})$, where $(x_{1},y_{1})$ is a point through which the line passes and $m$ is the slope. $(x_{1},y_{1})$ and a line perpendicular to the one we have to find are given. The equation given is in slope-intercept form, so it's evident that its slope is $-\frac{1}{2}$. Since the slopes of perpendicular lines are negative reciprocals, the slope of the line we have to find is: $m=-\dfrac{1}{\Big(-\dfrac{1}{2}\Big)}=2$ Now that $(x_{1},y_{1})$ and $m$ are known, substitute them into the point-slope form of the equation of a line formula and simplify to obtain the equation of the line we have to find: $y-y_{1}=m(x-x_{1})$ $y-2=2(x+3)$ $y-2=2x+6$ $y=2x+6+2$ $y=2x+8$, or, in general form, $2x-y+8=0$
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.