Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.2 Linear and Quadratic Functions - Exercises - Page 18: 17

Answer

The equation is: $5x-3y-1=0$.

Work Step by Step

The slope of the given line is $\frac{-3}{5}$, and thus the slope of the perpendicular line is $\frac{5}{3}$. Hence, we have a slope and a known point $(2,3)$. So, the equation of the perpendicular line is (using point-slope form of a line): $(y-y_1)=m(x-x_1)$ $(y-3) = \frac{5}{3} (x-2)$ and then $3y-9=5x-10$ Thus, the equation is $5x-3y-1=0$.
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