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: 5



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 } (1,3) \\\text{m}=-2 .\end{array} Use the properties of equality to express the equation in the standard form. $\bf{\text{Solution Details:}}$ Let $x_1=1,$ $y_1=3,$ and $m=-2.$ 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=-2(x-1) .\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-3=-2(x)-2(-1) \\\\ y-3=-2x+2 \\\\ 2x+y=2+3 \\\\ 2x+y=5 .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.