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: 18

Answer

$\text{a) Slope-Intercept Form: } y=2x-10 \\\text{b) Standard Form: } 2x-y=10$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the Slope Formula t of find the slope of the line passing through $(3,7)$ and $(5,6).$ Then use the Point-Slope Form of linear equations with the given point, $ (4,-2) $ 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= 7 ,\\y_2= 6 ,\\x_1= 3 ,\text{ and }\\ x_2= 5 .\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{7-6}{3-5} \\\\ m=\dfrac{1}{-2} \\\\ m=-\dfrac{1}{2} .\end{array} Since perpendicular lines have negative reciprocal slopes, then the needed line has the following characteristics: \begin{array}{l}\require{cancel} m=2 \\\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= 2 ,\end{array} is \begin{array}{l}\require{cancel} y-y_1=m(x-x_1) \\\\ y-(-2)=2(x-4) \\\\ y+2=2(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=2(x-4) \\\\ y+2=2(x)+2(-4) \\\\ y+2=2x-8 \\\\ y=2x-8-2 \\\\ y=2x-10 .\end{array} In the form $ax+by=c$ or the Standard Form, the equation above is equivalent to \begin{array}{l}\require{cancel} y=2x-10 \\\\ -2x+y=-10 \\\\ -1(-2x+y)=-1(-10) \\\\ 2x-y=10 .\end{array} Hence, $ \text{a) Slope-Intercept Form: } y=2x-10 \\\text{b) Standard Form: } 2x-y=10 .$
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