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: 17

Answer

$\text{a) Slope-Intercept Form: } y=\dfrac{2}{3}x+\dfrac{14}{3} \\\text{b) Standard Form: } 2x-3y=-14$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the Slope Formula t of find the slope of the line passing through $(3,9)$ and $(6,11).$ Then use the Point-Slope Form of linear equations with the given point, $ (-2,5) $ to find the equation of the needed line. Finally, Express the answer in both the Slope-Intercept and Standard forms. $\bf{\text{Solution Details:}}$ With the given points, \begin{array}{l}\require{cancel} y_1= 9 ,\\y_2= 11 ,\\x_1= 3 ,\text{ and }\\ x_2= 6 .\end{array} Using $m=\dfrac{y_1-y_2}{x_1-x_2}$ or the Slope Formula, the slope, $m,$ of the line is \begin{array}{l}\require{cancel} m=\dfrac{y_1-y_2}{x_1-x_2} \\\\ m=\dfrac{9-11}{3-6} \\\\ m=\dfrac{-2}{-3} \\\\ m=\dfrac{2}{3} .\end{array} Since parallel lines have same slopes, then the needed line has the following characteristics: \begin{array}{l}\require{cancel} m=\dfrac{2}{3} \\\text{Through: } (-4,2) .\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= 2 ,\\x_1= -4 ,\\m= \dfrac{2}{3} ,\end{array} is \begin{array}{l}\require{cancel} y-y_1=m(x-x_1) \\\\ y-2=\dfrac{2}{3}(x-(-4)) \\\\ y-2=\dfrac{2}{3}(x+4) .\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-2=\dfrac{2}{3}(x+4) \\\\ y-2=\dfrac{2}{3}(x)+\dfrac{2}{3}(4) \\\\ y-2=\dfrac{2}{3}x+\dfrac{8}{3} \\\\ y=\dfrac{2}{3}x+\dfrac{8}{3}+2 \\\\ y=\dfrac{2}{3}x+\dfrac{8}{3}+\dfrac{6}{3} \\\\ y=\dfrac{2}{3}x+\dfrac{14}{3} .\end{array} In the form $ax+by=c$ or the Standard Form, the equation above is equivalent to \begin{array}{l}\require{cancel} 3(y)=3\left( \dfrac{2}{3}x+\dfrac{14}{3} \right) \\\\ 3y=2x+14 \\\\ -2x+3y=14 \\\\ -1(-2x+3y)=-1(14) \\\\ 2x-3y=-14 .\end{array} Hence, $ \text{a) Slope-Intercept Form: } y=\dfrac{2}{3}x+\dfrac{14}{3} \\\text{b) Standard Form: } 2x-3y=-14 .$
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