Intermediate Algebra (12th Edition)

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

Chapter 2 - Chapter 2 Test: 12

Answer

$\text{a) Slope-Intercept Form: } y=-\dfrac{1}{2}x+2 \\\\\text{b) Standard Form: } x+2y=4$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the equation of the line with the given characeristics: \begin{array}{l}\require{cancel} \text{through } (-2,3) \text{ and } (6,-1) \end{array} use the Two-Point Form of linear equations. Express the answer in 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=-2 ,\\x_2=6 ,\\y_1=3 ,\\y_2=-1 ,\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-3=\dfrac{3-(-1)}{-2-6}(x-(-2)) \\\\ y-3=\dfrac{3+1}{-2-6}(x+2) \\\\ y-3=\dfrac{4}{-8}(x+2) \\\\ y-3=-\dfrac{1}{2}(x+2) .\end{array} In the form $y=mx+b$ or the Slope-Intercept Form, the equation above is equivalent to \begin{array}{l}\require{cancel} y-3=-\dfrac{1}{2}(x+2) \\\\ y-3=-\dfrac{1}{2}(x)-\dfrac{1}{2}(2) \\\\ y-3=-\dfrac{1}{2}x-1 \\\\ y=-\dfrac{1}{2}x-1+3 \\\\ y=-\dfrac{1}{2}x+2 .\end{array} In the form $ax+by=c$ or the Standard Form, the equation above is equivalent to \begin{array}{l}\require{cancel} y=-\dfrac{1}{2}x+2 \\\\ 2(y)=2\left( -\dfrac{1}{2}x+2 \right) \\\\ 2y=-x+4 \\\\ x+2y=4 .\end{array} Hence, the equation of the line is \begin{array}{l}\require{cancel} \text{a) Slope-Intercept Form: } y=-\dfrac{1}{2}x+2 \\\\\text{b) Standard Form: } x+2y=4 .\end{array}
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