Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.10 - Lines - 1.10 Exercises - Page 115: 47

Answer

$y=\dfrac{5}{2}x+\dfrac{1}{2}$, or, in general form, $5x-2y+1=0$

Work Step by Step

Through $(-1,-2);$ perpendicular to the line $2x+5y+8=0$ A point through which the line passes and the fact that the line is perpendicular to the line $2x+5y+8=0$ are known. Use the point-slope form of the equation of a line, which is $y-y_{1}=m(x-x_{1})$, where $(x_{1},y_{1})$ is a point through which the line passes and $m$ is the slope. Solve the equation of the given line for $y$. The resulting expression will be in the slope-intercept form and the slope can be identified as the number in front of $x$: $2x+5y+8=0$ $5y=-2x-8$ $y=-\dfrac{2}{5}x-\dfrac{8}{5}$ The slope of the given line is $-\dfrac{2}{5}$ Since the slopes of perpendicular lines are negative reciprocals, the slope of line whose equation must be found is: $m=-\dfrac{1}{\Big(-\dfrac{2}{5}\Big)}=\dfrac{5}{2}$ Substitute $(x_{1},y_{1})$ and $m$ into the point-slope form of the equation of a line formula and simplify to obtain the answer: $y-y_{1}=m(x-x_{1})$ $y+2=\dfrac{5}{2}(x+1)$ $y+2=\dfrac{5}{2}x+\dfrac{5}{2}$ $y=\dfrac{5}{2}x+\dfrac{5}{2}-2$ $y=\dfrac{5}{2}x+\dfrac{1}{2}$, or, in general form, $5x-2y+1=0$
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