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: 6

Answer

$x+y=6$

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 } (2,4) \\\text{m}=-1 .\end{array} Use the properties of equality to express the equation in the standard form. $\bf{\text{Solution Details:}}$ Let $x_1=2,$ $y_1=4,$ and $m=-1.$ 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-4=-1(x-2) .\end{array} Using the Distributive Property and 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} y-4=-1(x)-1(-2) \\\\ y-4=-x+2 \\\\ x+y=2+4 \\\\ x+y=6 .\end{array}
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