Question 332208: find the equation of the bisector of the acute angles and also the equation of the obtuse angles formed by the lines 7x-2y=4 and 3x+5y=15.
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! find the equation of the bisector of the acute angles and also the equation of the obtuse angles formed by the lines 7x-2y=4 and 3x+5y=15.
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Find the slope of each line:
7x-2y = 4, m = 7/2
3x+5y =15, m = -3/5
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The slope is the tangent of the angle with the x-axis, so the arctan(m) is the angle.
Add the 2 angles, and divide by 2 --> the angle of the bisector.
Angle of bisector = (arctan(7/2) + arctan(-3/5))/2 = 21.5454 degs
The slope is the tangent, m = 0.3948
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Find the intersection of the lines 7x-2y=4 and 3x+5y=15.
It's (50/41,93/41) or (1.2195,2.2683)
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Use the point and y = mx+b to find b
2.2683 = 0.3948*1.2195 + b
b = 1.7868
Equation of the bisector: y = 0.3948x + 1.7868
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I don't what you mean by "the equation of the obtuse angles"
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