Intermediate Algebra (12th Edition)

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

Chapter 2 - Section 2.3 - Writing Equations of Lines - 2.3 Exercises: 76

Answer

$\text{a) Slope-Intercept Form: } y=\dfrac{2}{5}x-\dfrac{39}{5} \\\text{b) Standard Form: } 2x-5y=39$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Find the slope of the line defined by the given equation, $ 5x+2y=18 .$ Then use the Point-Slope Form of linear equations with the given point, $ (2,-7) $ to find the equation of the needed line. Finally, Express the answer in both the Slope-Intercept and Standard forms. $\bf{\text{Solution Details:}}$ In the form $y=mx+b,$ the given equation is equivalent to \begin{array}{l}\require{cancel} 5x+2y=18 \\\\ 2y=-5x+18 \\\\ \dfrac{2y}{2}=\dfrac{-5x}{2}+\dfrac{18}{2} \\\\ y=-\dfrac{5}{2}x+9 .\end{array} Using $y=mx+b$ or the Slope-Intercept Form, where $m$ is the slope, then the slope of the line with the equation above is \begin{array}{l}\require{cancel} m=-\dfrac{5}{2} .\end{array} Since parallel lines have negative reciprocal slopes, then the needed line has the following characteristics: \begin{array}{l}\require{cancel} m=\dfrac{2}{5} \\\text{Through: } (2,-7) .\end{array} 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, \begin{array}{l}\require{cancel} y_1= -7 ,\\x_1= 2 ,\\m= \dfrac{2}{5} ,\end{array} is \begin{array}{l}\require{cancel} y-y_1=m(x-x_1) \\\\ y-(-7)=\dfrac{2}{5}(x-2) \\\\ y+7=\dfrac{2}{5}(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+7=\dfrac{2}{5}(x-2) \\\\ y+7=\dfrac{2}{5}(x)+\dfrac{2}{5}(-2) \\\\ y+7=\dfrac{2}{5}x-\dfrac{4}{5} \\\\ y=\dfrac{2}{5}x-\dfrac{4}{5}-7 \\\\ y=\dfrac{2}{5}x-\dfrac{4}{5}-\dfrac{35}{5} \\\\ y=\dfrac{2}{5}x-\dfrac{39}{5} .\end{array} In the form $ax+by=c$ or the Standard Form, the equation above is equivalent to \begin{array}{l}\require{cancel} 5(y)=5\left( \dfrac{2}{5}x-\dfrac{39}{5} \right) \\\\ 5y=2x-39 \\\\ -2x+5y=-39 \\\\ -1(-2x+5y)=-1(-39) \\\\ 2x-5y=39 .\end{array} Hence, $ \text{a) Slope-Intercept Form: } y=\dfrac{2}{5}x-\dfrac{39}{5} \\\text{b) Standard Form: } 2x-5y=39 .$
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