Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 4 Quadratic Functions and Factoring - 4.7 Complete the Square - 4.7 Exercises - Mixed Review - Page 291: 76


The equation of the line is $y=-\displaystyle \frac{2}{3}x+1$

Work Step by Step

$(x_{1},y_{1})=(3,-1)$ $(x_{2},y_{2})=(6,-3)$ Find the slope of the line: $m=\displaystyle \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ $m=\displaystyle \frac{-3-(-1)}{6-3}=-\frac{2}{3}$ You know the slope and a point on the line, so use point-slope form with either given point to write an equation of the line. Choose $(x_{1},y_{1})=(3,-1).$ $ y-y_{1}=m(x-x_{1})\qquad$ ...substitute $-1$ for $y_{1},\ -\displaystyle \frac{2}{3}$ for $m$ and $3$ for $x$. $ y-(-1)=-\displaystyle \frac{2}{3}(x-3)\qquad$ ...apply the Distributive Property. $ y+1=-\displaystyle \frac{2}{3}x+2\qquad$ ...add $-1$ to each side. $y=-\displaystyle \frac{2}{3}x+1$
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