Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 2 - Summary Exercises - Finding Slopes and Equations of Lines: 9

Answer

$\text{slope-intercept form: } y=-\dfrac{5}{6}x+\dfrac{13}{3} \\\\ \text{standard form: } 5x+6y=26$

Work Step by Step

Using $y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1)$ or the two-point form of linear equations, the equation of the line passing through $\left( -2,6 \right)$ and $\left( 4,1 \right)$ is \begin{array}{l}\require{cancel} y-6=\dfrac{6-1}{-2-4}\left( x-(-2) \right) \\\\ y-6=\dfrac{5}{-6}\left( x+2 \right) \\\\ y-6=-\dfrac{5}{6}\left( x+2 \right) .\end{array} In the form $y=mx+b$, the equation above is equivalent to \begin{array}{l}\require{cancel} y-6=-\dfrac{5}{6}x-\dfrac{5}{3} \\\\ y=-\dfrac{5}{6}x-\dfrac{5}{3}+6 \\\\ y=-\dfrac{5}{6}x-\dfrac{5}{3}+\dfrac{18}{3} \\\\ y=-\dfrac{5}{6}x+\dfrac{13}{3} .\end{array} In the form $Ax+By=C$, the equation above is equivalent to \begin{array}{l}\require{cancel} y=-\dfrac{5}{6}x+\dfrac{13}{3} \\\\ 6(y)=6\left( -\dfrac{5}{6}x+\dfrac{13}{3} \right) \\\\ 6y=-5x+26 \\\\ 5x+6y=26 .\end{array} Hence, the different forms of the equation of the line with the given conditions are \begin{array}{l}\require{cancel} \text{slope-intercept form: } y=-\dfrac{5}{6}x+\dfrac{13}{3} \\\\ \text{standard form: } 5x+6y=26 .\end{array}
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