Equation for the angle bisector of acute angle between two given lines..

I know there are two equations you get when you do this..
|distance from line1| = (+/-)|distance from line2|

the plus or minus gives you two equations...

My question is how do you know which one is for the acute angle and which one is for obtuse angle... such a question has been asked in bitsat it seems and i danot think u'll have enough time to calculate the angles using the tan inverse of the slopes...

Any hint wud be helpful... thankyou

1)
The equations of the bisectors of the angles between the lines
ax + by + c = 0 and a'x + b'y + c' = 0 (ab' ≠ a'b) are:
(ax + by + c) / √(a^2 + b^2) = ± (a'x + b'y + c') / √(a'^2 + b'^2).

2)
Proceed as follows to determine the acute angle bisector and obtuse angle bisector.
Write ax + by + c = 0 and a'x + b'y + c' = 0 such that constant terms are positive.
If aa' + bb' > 0, then the equation of the acute angle bisector is
(ax + by + c) / √(a^2 + b^2) = + (a'x + b'y + c') / √(a'^2 + b'^2) and the other equation is obtuse angle bisector.
If aa' + bb' < 0, then the equation of the acute angle bisector is
(ax + by + c) / √(a^2 + b^2) = - (a'x + b'y + c') / √(a'^2 + b'^2) and the other equation is obtuse angle bisector.

Edit:
For example, you want to find the acute angle bisector of the lines
3x + 4y + 10 = 0 and 5x - 12y - 20 = 0
First write both the equations in the form such that constant terms are positive.
=> 3x + 4y + 10 = 0 and - 5x + 12y + 20 = 0
Now check the sign of aa' + bb' which is 3 * (-5) + 4 * (12) > 0
=> acute angle bisector is
(3x + 4y + 10)/5 = + (5x - 12y - 20)/13
=> 14x + 112y + 230 = 0
=> 7x + 56y + 115 = 0