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 114: 27

Answer

$y=\dfrac{2}{3}x+\dfrac{19}{3}$ or, in general form, $2x-3y+19=0$

Work Step by Step

Through $(1,7);$ slope $\frac{2}{3}$ 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. Both $(x_{1},y_{1})$ and $m$ are given. Substitute them into the formula and simplify to obtain the equation of this line: $y-y_{1}=m(x-x_{1})$ $y-7=\dfrac{2}{3}(x-1)$ $y-7=\dfrac{2}{3}x-\dfrac{2}{3}$ $y=\dfrac{2}{3}x-\dfrac{2}{3}+7$ $y=\dfrac{2}{3}x+\dfrac{19}{3}$ or, in general form, $2x-3y+19=0$
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