Trigonometry: Urgent
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Trigonometry: Urgent

[From: ] [author: ] [Date: 12-07-05] [Hit: ]
they (likely) make an acute angle.Two, actually, one on each side of the intersection.If the origin (0,0) is contained within either of those acute angles,......
Here's the problem: Find an equation of the line that bisects the acute angle formed by the graphs of the equations 3x+y-9=0 and 2x-3y+96=0

I know the solution, and I know how to solve the problem;you use the distance from a point formula. However, I don't know how to determine the sign of the equations.

Here's how my text book puts it, and I don't quite understand it:
If the origin lies within the angle being bisected or the angle vertical to it, the distances from each line to a point on the bisector have the same sign. If the origin does not lie within the angle being bisected, the distances have opposite signs

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This is how I read it:

With the two lines drawn, they (likely) make an acute angle. Two, actually, one on each side of the intersection. If the origin (0,0) is contained within either of those acute angles, then the signs are the same.

But if the origin is not inside either /acute angle/, but instead is within either wider angle, then the signs are opposite to each other.

To put it another way -- let's assume that the angles the intersecting lines make are, say, 33 degrees and 147 degrees (and two more of the same).

If the point of origin is inside of one of the 33 degree angles, same sign.

If the point of origin is inside of one of the 147 degree angles, then, opposite signs.
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