<s>[INST] How to differentiate between external and internal angle bisectors of a triangle? I came up with this question when I was trying to figure out the coordinates of the incenter of a triangle with equations:
$4x-3y=0$, $3x-4y+12=0$, $3x+4y+2=0$.
I assumed the coordinates of the incenter to be $(h,k)$ and equated the perpendicular distances from all the sides and got:
$4h-3k=\pm(3h-4k+12)$, $3h-4k+12=\pm(3h+4k+2)$, $3h+4k+2=\pm(4h-3k) $
But I didn't know which sign to take. As different signs would mean different angle bisectors(external or internal). 
Image showing intersection of Angle bisectors(posted as a link due to low reputation restrictions)
I found this Can we find incentre of a triangle by using equation of lines?, answering which bisector to take but I couldn't understand the method in the case when I do not have the vertices.
P.S. I want to know the incenter without knowing the vertices. Also I am curious why the above method works, intuitively. Thanks.
 [/INST] $4·x-3·y=0$
$3·x-4·y+12=0$
$3·x+4·y+2=0$
multiply the equations by the signs of the cofactors of the constants: 0, 12 and 2.
$M=\left [ \begin{array}{} 4 & -3 & 0 \\ 3 & -4 & 12 \\ 3 & 4 & 2\\ \end{array} \right ] $
$cofactor{M_{13}}=\left|\begin{array}{} 3 & -4 \\ 3 & 4 \\ \end{array}   \right| =24$
$cofactor{M_{23}}=-\left|\begin{array}{} 4 & -3 \\ 3 & 4 \\ \end{array}   \right| =-25$
$cofactor{M_{33}}=\left|\begin{array}{} 4 & -3 \\ 3 & -4 \\ \end{array}   \right| =-7$
sign(24)=1, sign(-25)=-1 and sign(-7)=-1
$1·(4·x-3·y)=0⇒4·x-3·y=0$
$-1·(3·x-4·y+12)=0⇒-3·x+4·y-12=0$
$-1·(3·x+4·y+2)=0⇒-3·x-4·y-2=0$
with these "oriented" equations we obtain the bisectors
(the three generate internal bisectors)
$\frac{4·x-3·y}{\sqrt{4^2+(-3)^2}}=\frac{-3·x-4·y-2}{\sqrt{(-3)^2+(-4)^2}}$
$\frac{-3·x+4·y-12}{\sqrt{(-3)^2+4^2}}=\frac{-3·x-4·y-2}{\sqrt{(-3)^2+(-4)^2}}$
$\frac{4·x-3·y}{\sqrt{4^2+(-3)^2}}=\frac{-3·x+4·y-12}{\sqrt{(-3)^2+4^2}}$
$7·x+y+2=0$
$4·y-5=0$
$7·x-7·y+12=0$
bisectorInternal
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