Intermediate Algebra (12th Edition)

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

Chapter 2 - Chapter 2 Test: 5

Answer

$y=\dfrac{1}{2}x+1$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the Two-Point Form of linear equations to find the equation of the line passing through $( 6,4)$ and $(-4,-1).$ $\bf{\text{Solution Details:}}$ Using $y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1)$ or the Two-Point Form of linear equations, where \begin{array}{l}\require{cancel} x_1=6 ,\\x_2=-4 ,\\y_1=4 ,\\y_2=-1 ,\end{array} the equation of the line is \begin{array}{l}\require{cancel} y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1) \\\\ y-4=\dfrac{4-(-1)}{6-(-4)}(x-6) \\\\ y-4=\dfrac{4+1}{6+4}(x-6) \\\\ y-4=\dfrac{5}{10}(x-6) \\\\ y-4=\dfrac{1}{2}(x-6) \\\\ y-4=\dfrac{1}{2}(x)+\dfrac{1}{2}(-6) \\\\ y-4=\dfrac{1}{2}x-3 \\\\ y=\dfrac{1}{2}x-3+4 \\\\ y=\dfrac{1}{2}x+1 .\end{array}
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