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Question 1011461: Find the equation of the angle bisector of the acute angles formed by the lines x+3y=9 and 4x+y=8.
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Find the equation of the angle bisector of the acute angles formed by the lines x+3y=9 and 4x+y=8.
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Find the intersection of the 2 lines.
x +3y=9
12x+3y=24 2nd eqn times 3
----------------- Subtract
-11x = -15
x = 15/11
y = 3 - x/3 = 28/11
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--> (15/11,28/11) is the intersection
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Use two lines thru the Origin with the same slopes to find the slope of the bisector.
x+3y=9 --> m1 = -1/3
4x+y=8.--> m2 = -4
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The lines thru the Origin are:
y = -x/3 and y = -4x
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The 2 slopes are -1/3 and -4
The slope is the tangent of the angle with the x-axis.
atan(-1/3) =~ -18.4349 degs
atan(-4) =~ -75.9638 degs
The average =~ -47.1993 degs
The slope of the bisector = tan(-47.1993) = -1.08
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Use y-y1 = m*(x-x1)
y - 28/11 = -1.08*(x - 15/11)
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