given two lines in 2D, how to select the angle bisector related to the smallest angle between the lines

I have two lines:

first line: $a_1x+b_1y=c_1 \qquad(1)$

second line: $a_2x+b_2y=c_2 \qquad(2)$

I know that the two angle bisectors are expressed by

$\frac{a_{1}x+b_{1}y-c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}=\pm \frac{a_{2}x+b_{2}y-c_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}}}\qquad (3)$

Is there any link between the sign of RHS in (3) and the bisector of the smallest (biggest) angle?

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I don't think you can make any such connections. Maybe if you know the relative positions of the two lines. –  Beni Bogosel May 30 '11 at 14:37
The plus sign on the RHS of $(1)$ may correspond either to the smaller or to the bigger angle between the two lines, as illustrated by the following examples 1) and 2). 1) For the system $$x+y=0,\qquad 2x-y=0$$ the angle bisector whose equation takes the plus sign $$\frac{x+y}{\sqrt{1+1}}=+\frac{2x-y}{\sqrt{4+1}}$$ corresponds to the bigger angle. 2) For the system $$y=0,\qquad x-y=0$$ the angle bisector whose equation takes the plus sign $$\frac{y}{\sqrt{1}}=+\frac{x-y}{\sqrt{1+1}}$$ corresponds to the smaller angle. –  Américo Tavares May 31 '11 at 7:51

Forget about $c_1$ and $c_2$, put $u:=(a_1,b_1)/\sqrt{a_1^2+b_1^2}$, $v:=(a_2,b_2)/\sqrt{a_2^2+b_2^2}$ and let $z:=(x,y)$. The lines $u\cdot z=0$ and $v\cdot z =0$ are parallel to your lines $g_1$ and $g_2$. If $u\cdot v>0$ (i.e., $u$ and $v$ enclose an acute angle) then it easy to see that $u+v$ is orthogonal to the bisector of the smaller angle between $g_1$ and $g_2$, so this bisector is parallel to the line $(u+v)\cdot z=0$. If, on the other hand, $u\cdot v<0$ then $u$ and $-v$ enclose an acute angle; therefore $u-v$ is orthogonal to the bisector of the smaller angle between $g_1$ and $g_2$, so in this case the desired bisector is parallel to the line $(u-v)\cdot z=0$.

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While it's a substantially different formulation of the answer (and uses a different formulation of the given lines), if you were to use my answer here on finding equations of the angle bisectors using trigonometry, you could tell from the rotational angles of the two given lines ($\theta_1=\arctan(m_1)$ and $\theta_2=\arctan(m_2)$) whether the desired angle bisector had rotational angle $\frac{\theta_1+\theta_2}{2}$ (if $|\theta_1-\theta_2|\le\frac{\pi}{2}$) or $\frac{\pi+\theta_1+\theta_2}{2}$ (if $|\theta_1-\theta_2|\ge\frac{\pi}{2}$).