Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Appendices - Section A.3 - Lines, Circles, and Parabolas - Exercises A.3 - Page AP-17: 15


$y=-\displaystyle \frac{1}{2}x+12$

Work Step by Step

We solve the given equation for y (write it in slope-intercept form) $6x-3y=5$ $-3y=-6x+5$ $y=\displaystyle \frac{-6}{-3}x+\frac{5}{-3}$ $y=2x-\displaystyle \frac{5}{3}$ The slope of the line is $m_{1}=2$, and any line perpendicular to it has slope $m_{2}$ such that $m_{2}=-\displaystyle \frac{1}{m_{1}}$ $m_{2}=-\displaystyle \frac{1}{2}$ Now, find the perpendicular line through $(4,10)$. The point-slope equation of a line containing the point $(x_{1},y_{1})$, with slope $m$ is $y=y_{1}+m(x-x_{1})$ Given $(x_{1},y_{1})=(4,10)$ and $m=-\displaystyle \frac{1}{2}$ $y=10+(-\displaystyle \frac{1}{2})(x-4)$ $y=10-\displaystyle \frac{1}{2}x+2$ $y=-\displaystyle \frac{1}{2}x+12$
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