In the cartesian coordinate plane,
Let $f_k(x,y) = a_kx + b_ky + c_k$.
Let $f_1(x,y)=0$ and $f_2(x,y)=0$ be two lines,
We know their (2) angle bisectors are given by :
$\frac{f_1(x,y)}{\sqrt{a_1^2+b_1^2}} = \pm \frac{f_2(x,y)}{\sqrt{a_2^2+b_2^2}}$
Now, the question is which one is acute and which is obtuse angle bisector, Say one bisector is $B_1$ and one is $B_2$, one way to do this would be to find angle between the bisector and one line, is the modulus of the tangent of that angle is less than $1$, that is acute and otherwise, obtuse.
Another way which was given in my book was;
Let sign of $a_1a_2+b_1b_2$ be $S$, ie. $S$ is $\frac{a_1a_2+b_1b_2}{|a_1a_2+b_1b_2|}$, then the obtuse angle bisector is given by :
$\frac{f_1(x,y)}{\sqrt{a_1^2+b_1^2}} = S \frac{f_2(x,y)}{\sqrt{a_2^2+b_2^2}} \ \ \ (^@)$
Now, this is something I was just not able to prove, like I tried to do to use the tangent less than $1$ condition but it just did not work at all, at least for me it became really ugly really fast (as I had to square it to take care of modulus and signs).
But, I do think there could be a solution using VECTORS!!!
The reason I believe so is that if $\vec c = a_1 \hat i + b_1 \hat j$ and $\vec d = a_2 \hat i + b_2 \hat j$,
Then, $\vec c \cdot \vec d = a_1a_2 + b_1b_2$.
But I haven't really learnt vectors much, so I am not really sure how they would relate to a straight line.
So, in a nutshell, I just want the vector proof of $(^@)$