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Question 1200254: Find the equation of the line through the point (5,0) and perpendicular to the line y=(4x/3)+5. Express the equation of the line in slope-intercept form, y=mx+b.
So I am assuming that I need to put the points (5,0) into the y=mx+b so I did that. 0=(4/3)(5/1)+b and solved.
0=(20/3)+b so then I solved again so is b=-(20/3)?
I don't know what part of this I am doing wrong but I want to make sure I understand the content. Any help is greatly appreciated :)
Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Find the equation of the line through the point (5,0) and perpendicular to the line y=(4x/3)+5. Express the equation of the line in slope-intercept form, y=mx+b.
So I am assuming that I need to put the points (5,0) into the y=mx+b so I did that. 0=(4/3)(5/1)+b and solved.
0=(20/3)+b so then I solved again so is b=-(20/3)?
I don't know what part of this I am doing wrong but I want to make sure I understand the content. Any help is greatly appreciated :)
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The slope of lines perpendicular to the given line is the negative inverse, -3/4.
Then, y-y1 = m*(x-x1) where (x1,y1) is the given point.
y - 0 = (-3/4)*(x - 5)
y = (-3/4)x + 15/4
Answer by ikleyn(52754)  (Show Source):
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