College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.3 - Lines - 1.3 Exercises - Page 113: 47


$y=\dfrac{5}{2}x + \dfrac{1}{2}$

Work Step by Step

RECALL: (1) The slope-intercept form of a line's equation is $y=mx+b$ where m = slope and b = y-intercept. (2) Perpendicular lines have slopes that are negative reciprocals of each other (product is $-1$). Transform the line $2x+5y+8-0$ to slope-intercept form to obtain: $5y=-2x-8 \\\dfrac{5y}{5} = \dfrac{-2x-8}{5} \\y=-\dfrac{2}{5}x - \dfrac{8}{5}$ The slope of this line is $-\dfrac{2}{5}$. The line we are looking for the equation of is perpendicular to the line above. This means that its slope is the negative reciprocal of $-\dfrac{2}{5}$, which is $\dfrac{5}{2}$. Thus, the tentative equation of the line we are looking for is: $y=\dfrac{5}{2}x+b$ The line passes through the point $(-1, -2)$. To find the value of $b$, substitute the x and y coordinates of this point into the tentative equation above to obtain: $y=\dfrac{5}{2}x + b \\-2 = \dfrac{5}{2}(-1) + b \\-2 = -\dfrac{5}{2} + b \\-2+\dfrac{5}{2}=b \\-\dfrac{4}{2} + \dfrac{5}{2} = b \\\dfrac{1}{2} = b$ Therefore, the equation of the line we are looking for is: $y=\dfrac{5}{2}x + \dfrac{1}{2}$
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