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 - 2.5 Exercises - Page 226: 14



Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the Two-Point Form of linear equations to find the equation of the line with the following given characteristcs: \begin{array}{l}\require{cancel} \text{through } (2,3) \text{ and } (-1,2) .\end{array} Use the properties of equality to express the equation in the standard form. $\bf{\text{Solution Details:}}$ 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=2 ,\\x_2=-1 ,\\y_1=3 ,\\y_2=2 ,\end{array} the equation of the line is \begin{array}{l}\require{cancel} y-3=\dfrac{3-2}{2-(-1)}(x-2) \\\\ y-3=\dfrac{3-2}{2+1}(x-2) \\\\ y-3=\dfrac{1}{3}(x-2) .\end{array} Using the properties of equality, in the form $ax+by=c$ or the Standard Form, the equation above is equivalent to \begin{array}{l}\require{cancel} 3(y-3)=\left[ \dfrac{1}{3}(x-2) \right]3 \\\\ 3(y-3)=1(x-2) \\\\ 3y-9=x-2 \\\\ -x+3y=-2+9 \\\\ -x+3y=7 \\\\ -1(-x+3y)=-1(7) \\\\ x-3y=-7 .\end{array}
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