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: 80


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

Work Step by Step

$(x_{1},y_{1})=(6,3)$ $(x_{2},y_{2})=(8,4)$ Find the slope of the line: $m=\displaystyle \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ $m=\displaystyle \frac{4-3}{8-6}=\frac{1}{2}$ 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 $(x_{1},y_{1})=(6,3).$ $ y-y_{1}=m(x-x_{1})\qquad$ ...substitute $3$ for $y_{1},\ \displaystyle \frac{1}{2}$ for $m$ and $6$ for $x$. $ y-3=\displaystyle \frac{1}{2}(x-6)\qquad$ ...apply the Distributive Property. $ y-3=\displaystyle \frac{1}{2}x-3\qquad$ ...add $3$ to each side. $y=\displaystyle \frac{1}{2}x$
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