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

Question

I have two lines:

$$a_1x+b_1y=c_1 \tag{1}$$ $$a_2x+b_2y=c_2 \tag{2}$$

I know that the two angle bisectors are expressed by

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

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

Answer 1

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 $(u-v)\cdot z=0$.

Answer 2

The $+/-$ have to do with the fact that of the two angle bisectors one will be such that the points (of which it is the locus) will result in distances (with same sign say positive) from the two lines and the other angle bisector (which is perpendicular to the first angle bisector) will therefore result in distances of opposite sign from the two lines.

So I don't think we can say which choice of sign will give you the bisector of the bigger angle. For that you may find out the slope of the angle bisector of the two lines and compare with the ones you obtain by choosing the appropriate sign in the two equations above.

Answer 3

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 $|\theta_1-\theta_2|\ge\frac{\pi}{2}$).

Of course, this is nowhere near as nice as what you were hoping/looking for.

Answer 4

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

No, there is not.

Suppose there is - for example let the "+" sign determine the bisector of the smallest angle, so its equation is

$$\frac{a_{1}x+b_{1}y-c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}= \color{red}+ \frac{a_{2}x+b_{2}y-c_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}}}\tag{3'}$$

Now, let's rewrite your (original) equation for the second line

$$a_2x+b_2y=c_2 \tag{2}$$

as

$$(-a_2)x+(-b_2y)=-c_2 \tag{2'}$$

This is an equation of the same line. Substituting its coefficients ($-a_2, -b_2, -c_2$) into $(3')$ we obtain

$$\frac{a_{1}x+b_{1}y-c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}= \color{red} - \frac{a_{2}x+b_{2}y-c_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}}}\tag{3''}$$

Oops! $(3')$ and $(3'')$ for the bisector of the smallest angle - and for the same lines!

(The point is that equations $(1)$ and $(2)$ are not canonical - there are many such equations for the same line.)

Answer 5

If $c_1, c_2$ are positive then the angle bisector of the acute angle between the lines $L_1:a_1x+b_1y+c_1=0$ and $L_2:a_2x+b_2y+c_2=0$ is

$B_1: \frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}=\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$, if $a_1a_2+b_1b_2\lt0$

$B_2: \frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}=-\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$, if $a_1a_2+b_1b_2\gt0$

Proof

The $\tan$ of angle between first line $L_1$ and bisector $B_1$ can be worked out as $$|\tan \theta| =\frac{|p|}{\sqrt{p^2+q^2-q}}, \quad p=a_1b_2-b_1a_2, \quad q=a_1a_2+b_1b_2.$$ When $q<0 \implies |\tan \theta|<1 \implies \theta < \pi/4.$ So $B_1$ given by you is acute angle bisector.

Next, if $q>0$, $$\sqrt{p^2+q^2}\le |p|+|q| \implies \sqrt{p^2+q^2}-q\le |p|+|q|-q=|p| \implies |\tan \theta|>1.$$ Then $B_1$ will be obtuse angle bisector.