Find the equation of the line passing through (6,6) and perpendicular to the 5x-3y-8=0
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# Find the equation of the line passing through (6,6) and perpendicular to the 5x-3y-8=0

[From: ] [author: ] [Date: 11-09-26] [Hit: ]
and -1/5 being the slope. Hope Im right.......
First, find the slope of the line defined by 5x-3y-8=0
5x - 3y - 8 = 0 add 8 to both sides
5x - 3y = 8 subtract 5x from both sides
-3y = -5x + 8 divide both sides by -3
y = (5/3)x - 8/3 given that y = mx + b, where m is the slope, m = 5/3

a line perpendicular to this line has the negative inverse slope or, in this case, -3/5

use point slope to define your new line, (y - y1) = m(x - x1) with (x1,y1) = (6,6)
(y - 6) = (-3/5)(x - 6) distribute the -3/5
y - 6 = (-3/5)x + 18/5 add 6 to both sides
y = (-3/5)x + 48/5 ANSWER

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you are welcome, Fiona was on the right track but she had to get the equation into slope-intercept form before determining the slope!!!!

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The slope of that line is 5, and the slope of a perpendicular line is its negative reciprocal. So the equation would be y-6= -1/5 (x-6) in point slope from. In slope intercept it would be y=1/5x +4.8, with 4.8 being the y intercept, and -1/5 being the slope. Hope I'm right.
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