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: 8



Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the different forms of linear equations to find the equation of the line with the following given characteristcs: \begin{array}{l}\require{cancel} \text{through } (-4,3) \\\text{m}=\dfrac{3}{4} .\end{array} Use the properties of equality to express the equation in the standard form. $\bf{\text{Solution Details:}}$ Let $x_1=-4,$ $y_1=3,$ and $m=\dfrac{3}{4}.$ Using $y-y_1=m(x-x_1)$ or the Point-Slope Form of linear equations, the equation of the line with the given conditions is \begin{array}{l}\require{cancel} y-3=\dfrac{3}{4}(x-(-4)) \\\\ y-3=\dfrac{3}{4}(x+4) .\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{3}{4}(x+4) \right]4 \\\\ 4(y-3)=3(x+4) \\\\ 4(y)+4(-3)=3(x)+3(4) \\\\ 4y-12=3x+12 \\\\ -3x+4y=12+12 \\\\ -3x+4y=24 \\\\ -1(-3x+4y)=-1(24) \\\\ 3x-4y=-24 .\end{array}
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