SOLUTION: Find the equation of the bisector of the pair of acute angles formed by the lines 4x-3y= 8 and 2x+y=4
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Question 374054: Find the equation of the bisector of the pair of acute angles formed by the lines 4x-3y= 8 and 2x+y=4
Found 2 solutions by melledwyer, Alan3354:
Answer by melledwyer(2) (Show Source): You can put this solution on YOUR website!
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Find the intersection of the 2 lines, = (2,0)
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Find the slope of the 2 lines:
4x-3y = 8 m1 = 4/3
2x+y = 4 m2 = -2
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The slope is the arctan of the angle the line makes with the x-axis
arctan(4/3) =~ 53.13º
arctan(-2) =~ -63.435º
The angle between them is 116.565º (in Quad 1), so the acute angles are 83.435º
The bisector makes and angle of 53.15 + (83.435)/2 = 84.8675º with the x-axis
Its slope = tan(84.8675) = 11.133
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Use y = mx + b and the point (2,0)
0 = 2*11.133 + b
b = -22.266
Equation is y = 11.133x - 22.266
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