Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 2 - Graphs and Functions - 2.5 Equations of Lines and Linear Models - 2.5 Exercises: 21

Answer

$\color{blue}{y=\frac{2}{3}(x-3)}$

Work Step by Step

RECALL: (1) The point-slope form of a line's equation is $y-y_1=m(x-x_1)$ where $m$=slope and $(x_1, y_1)$ is a point on the line. (2) The slope $m$ of a line that contains the points $(x_1, y_2)$ and $(x_2, y_2)$ is given by the formula $m=\dfrac{y_2-y_1}{x_2-x_1}$.. Solve for the slope using the formula in (2) above to obtain: $m=\dfrac{-2-0}{0-3} \\m=\dfrac{-2}{-3} \\m=\dfrac{2}{3}$ With a slope of $\frac{2}{3}$, either of the two given points can be used to write the point-slope form of the line's equation. Therefore, using the point $(3, 0)$ gives: $y-0 = \frac{2}{3}(x-3) \\\color{blue}{y=\frac{2}{3}(x-3)}$
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