Intermediate Algebra (12th Edition)

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

Chapter 2 - Summary Exercises - Finding Slopes and Equations of Lines: 15

Answer

$\text{a) Slope-Intercept Form: } y=-\dfrac{5}{2}x \\\text{b) Standard Form: } 5x+2y=0$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Find the slope of the line defined by the given equation, $ 2x-5y=6 .$ Then use the Point-Slope Form of linear equations with the given point, $ (0,0) $ 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} 2x-5y=6 \\\\ -5y=-2x+6 \\\\ \dfrac{-5y}{-5}=\dfrac{-2x}{-5}+\dfrac{6}{-5} \\\\ y=\dfrac{2}{5}x-\dfrac{6}{5} .\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{2}{5} .\end{array} Since perpendicular lines have negative reciprocal slopes, then the needed line has the following characteristics: \begin{array}{l}\require{cancel} m=-\dfrac{5}{2} \\\text{Through: } (0,0) .\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= 0 ,\\x_1= 0 ,\\m= -\dfrac{5}{2} ,\end{array} is \begin{array}{l}\require{cancel} y-y_1=m(x-x_1) \\\\ y-0=-\dfrac{5}{2}(x-0) \\\\ y=-\dfrac{5}{2}x .\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=-\dfrac{5}{2}x .\end{array} In the form $ax+by=c$ or the Standard Form, the equation above is equivalent to \begin{array}{l}\require{cancel} 2(y)=2\left( -\dfrac{5}{2}x \right) \\\\ 2y=-5x \\\\ 5x+2y=0 .\end{array} Hence, $ \text{a) Slope-Intercept Form: } y=-\dfrac{5}{2}x \\\text{b) Standard Form: } 5x+2y=0 .$
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