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Question 368737: I'm stumped!
Write an equation for the line perpendicukar to MN that contains point P.
MN:y=2/3x, P(6,6)
Slope of given the given line is 2/3
so than the book tells me to put it into a equation like:
2/3m=-1
I get that but how do I get 2/3 to the other side
2/3m=-1
I think I divide the 2/3 on both side to get rid of it on the m side but how do I multiply 2/3 by -1?
I am following the instructions in the book but thats what I'm struggling with!
Sorry this is confusing!!!
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Write an equation for the line perpendicukar to MN that contains point P.
MN:y=2/3x, P(6,6)
Slope of given the given line is 2/3
-------------------
The slope of lines perpendicular is the negative inverse, = -3/2
Use y = mx + b and the point to find b, the y-intercept.
6 = (-3/2)*6 + b
b = 15
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--> y = (-3/2)x + 15
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so than the book tells me to put it into a equation like:
2/3m=-1
I get that but how do I get 2/3 to the other side
2/3m=-1
I think I divide the 2/3 on both side to get rid of it on the m side but how do I multiply 2/3 by -1?
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