Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Appendix A - Appendix A Exercises - Page 1158: 43

Answer

$$y = \frac{1}{3}x + 7$$

Work Step by Step

$$\eqalign{ & {\text{Passing through }}\left( { - 3,6} \right) \cr & \underbrace {y = - 3x + 2}_{y = mx + b}{\text{ }}\left( {\bf{1}} \right) \cr & {\text{Two lines with slopes }}{m_1}{\text{ and }}{m_2}{\text{ are perpendicular if and only}} \cr & {\text{if their slopes are negative reciprocals}} \cr & {m_2} = - \frac{1}{{{m_1}}} \cr & {\text{From }}\left( {\bf{1}} \right){\text{ let }}{m_1} = - 3,{\text{ then}} \cr & {m_2} = - \frac{1}{{{m_1}}} = - \frac{1}{{ - 3}} \cr & {m_2} = \frac{1}{3} \cr & {\text{The line passes through the point }}\underbrace {\left( { - 3,6} \right)}_{\left( {{x_1},{y_1}} \right)} \cr & {\text{Use the point - slope form of the equation of a line}} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 6 = \frac{1}{3}\left( {x + 3} \right) \cr & y - 6 = \frac{1}{3}x + 1 \cr & y = \frac{1}{3}x + 7 \cr} $$
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