# Regarding an answer on angular bisectors in 2D coordinate geometry.

I am aware that the following expression represents the two angular bisectors for two straight lines.

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

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

I then had the following questions regarding the best answer:

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

1. What are $$u$$ and $$v$$ and $$z$$? Hence what would $$u+v$$ and $$u-v$$ be? (I am unable to understand the notation.)
2. How are the dot products parallel to the original lines? (Probably answerable from 1.)
3. Is there a way to use the original equations to tell if a bisector is 'internal' or 'external'?

Apologies for the silly question and for violating any rules. I do not have enough reputation to ask in the comments.

• Re: 3: the sign of $u\cdot v$ is equal to the sign of $(a_1,b_1)\cdot(a_2,b_2)$.
– amd
Commented Mar 12, 2019 at 6:05

Ad $$1$$:

$$u, v$$ are unit vectors perpendicular to those lines. $$z := (x, y)$$ is a $$formal$$ vector, so for example

$$u \cdot z = 0$$

is an equation of a line, because it is

$$(u_1, u_2) \cdot(x,y) = 0$$ or $$u_1x+u_2y=0$$ $$(u+v)$$ and $$(u-v)$$ are perpendicular to bisectors $$f, g$$ of original lines $$m,n$$:

Ad $$2$$:

As you guessed, it follows from the previous part.

Ad $$3$$:

Yes, there is. Follow the link in your own question to learn how.

• The diagram does help tremendously as I am trying to work my head around 1 and 2. But how does Ad.3 tell us which equation refers to the 'internal' bisector and which to the 'external'? So when '$+$' is used you get one of the bisectors and the other is obtained when '$-$' is used, but which one bisects the acute and which one the obtuse? Commented Mar 11, 2019 at 13:03
• It doesn't tell us, I showed that it is impossible. The problem is that the same line has (infinitely) many different equations. Commented Mar 11, 2019 at 13:10
• Regarding 1 and 2. : Since $u,v$ are perpendicular to $m,n$ thus the vector $u+v$ (which is the acute bisector for $u,v$) is the 'obtuse' bisector for the original pair of lines. Thus $(u+v)\cdot z=0$ which is perpendicular to $u+v$ is the 'acute bisector' for the original lines?! Sorry for being painfully slow but why is $u\cdot z = 0$ parallel to the original? Commented Mar 11, 2019 at 13:29
• See, you put a question and I answered it. You didn't accept my answer. It seems that you are not satisfied with it, so you have wait for a better one. Commented Mar 11, 2019 at 15:04
• Sorry for being a bit dense but can I ask follow up questions? Commented Mar 11, 2019 at 16:44