Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 2 - Review Exercises: 32

Answer

$\text{a) Slope-Intercept Form: } y=-x+2 \\\\\text{b) Standard Form: } x+y=2$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the equation of the line that passes through the points $( 0,2 )$ and $( 2,0 ),$ (see Exercise 18) use the Two-Point Form of linear equations. Give the equation in the Slope-Intercept Form and 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=0 ,\\x_2=2 ,\\y_1=2 ,\\y_2=0 ,\end{array} the equation of the line is \begin{array}{l}\require{cancel} y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1) \\\\ y-2=\dfrac{2-0}{0-2}(x-0) \\\\ y-2=\dfrac{2}{-2}(x) \\\\ y-2=-1(x) \\\\ y-2=-x .\end{array} Using the properties of equality, in the form $y=mx+b$ or the Slope-Intercept Form, the equation above is equivalent to \begin{array}{l}\require{cancel} y-2=-x \\\\ y=-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} y=-x+2 \\\\ x+y=2 .\end{array} The equation of the line is \begin{array}{l}\require{cancel} \text{a) Slope-Intercept Form: } y=-x+2 \\\\\text{b) Standard Form: } x+y=2 .\end{array}
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