College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 2 - Section 2.5 - Equations of Lines and Linear Models - Summary Exercises on Graphs, Circles, Functions, and Equations - Page 231: 9

Answer

$y=-\frac{1}{3}x+\frac{1}{3}$ See graph.

Work Step by Step

To calculate the slope between points $(x_1,y_1)$ and $(x_2,y_2)$, we use the formula: $slope=m=\frac{y_2-y_1}{x_2-x_1}$ We calculate the slope between $(-2,1)$ and $(4,-1)$: $slope=\displaystyle \frac{-1-1}{4-(-2)}$ $=\frac{-2}{6}$ $=-\frac{1}{3}$ A line in point-slope form has the equation: $y-y_1=m(x-x_1)$ We plug in the point $(4,-1)$: $y-(-1)=-\displaystyle \frac{1}{3}(x-4)$ And solve for $y$: $y+1=-\displaystyle \frac{1}{3}(x-4)$ $3(y+1)=-(x-4)$ $3y+3=-x+4$ $3y=-x+1$ $y=-\frac{1}{3}x+\frac{1}{3}$ See graph.
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