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: 13



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 } (-1,3) \text{ and } (3,4) .\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=-1 ,\\x_2=3 ,\\y_1=3 ,\\y_2=4 ,\end{array} the equation of the line is \begin{array}{l}\require{cancel} y-3=\dfrac{3-4}{-1-3}(x-(-1)) \\\\ y-3=\dfrac{-1}{-4}(x+1) \\\\ y-3=\dfrac{1}{4}(x+1) .\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} 4(y-3)=\left[\dfrac{1}{4}(x+1)\right]4 \\\\ 4(y-3)=1(x+1) \\\\ 4(y)+4(-3)=1(x)+1(1) \\\\ 4y-12=x+1 \\\\ -x+4y=1+12 \\\\ -x+4y=13 \\\\ -1(-x+4y)=-1(13) \\\\ x-4y=-13 .\end{array}
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