<s>[INST] Can we find incentre of a triangle by using equation of lines? My question is, in the manner which we find the orthocentre  of a triangle merely by using it's side's line equations, can we also find the incentre? 
For example, consider,
$$3x+4y-7=0$$
$$4x-3y+19=0$$
$$18x-6y+7=0$$
We can find the orthocentre of the triangle made by these lines using the fact that in a right angled triangle, the vertex containing the right angle is the orthocentre.

So, for any triangle made by three straight lines, can we find the incentre as well?

I will appreciate both short and long answers. Any thoughts are welcome. Thanks!
EDIT :I will appreciate if someone gives me a method where I don't have to solve for the vertices.
 [/INST] The intersection point (Incenter) of the internal bisectors can be obtained through a formula with the cofactors, coefficients and constants of the equations.
$A·x+B·y+C=0$
$D·x+E·y+F=0$
$G·x+H·y+I=0$
$M=\left( \begin{array}{} A & B & C \\ D & E & F \\ G & H & I \\ \end{array}  \right) $
$p=\sqrt{A^2+B^2}$
$q=\sqrt{D^2+E^2}$
$r=\sqrt{G^2+H^2}$
$xI=\frac{(\frac{p·cA }{ sgn(cC)} +\frac{q· cD}{sgn(cF)}+ \frac{r ·cG}{sgn(cI)})}{p·|cC| + q· |cF| + r ·|cI|}$
$yI=\frac{(\frac{p·cB }{ sgn(cC)} +\frac{q· cE}{sgn(cH)}+ \frac{r ·cG}{sgn(cI)})}{p·|cC| + q· |cF| + r ·|cI|}$
where cA, cB, cC ... cI are the cofactors of the matrix M
calculating the Incenter with its equations
$M=\left( \begin{array}{} -3 & 8 & -6 \\ 4 & -4 & 14 \\ 18 & -6 & 7 \\ \end{array}  \right)$
$CofactorsM=\left( \begin{array}{} 56 & 224 & 48 \\ -20 & 87 & 126 \\ 88 & 18 & -20 \\ \end{array}  \right) $
$p=\sqrt{(-3)^2+8^2}=\sqrt{73}$
$q=\sqrt{4^2+(-4)^2}=\sqrt{32}$
$r=\sqrt{18^2+(-6)^2}=\sqrt{360}$
$xI=\frac{\frac{\sqrt{73}·56}{1}+\frac{\sqrt{32}·(-20)}{1}+\frac{\sqrt{360}·88}{-1}}{\sqrt{73}·|48|+\sqrt{32}·|126|+\sqrt{360}·|-20|}=-0.868$
$ yI=\frac{\frac{\sqrt{73}·(224)}{1}+\frac{\sqrt{32}·(87)}{1}+\frac{\sqrt{360}·(18)}{-1}}{\sqrt{73}·|48|+\sqrt{32}·|126|+\sqrt{360}·|-20|}=1.374$
alternative formula:
$xI2=\frac{\left| \left ( \begin{array}{} cA \\ cD \\ cG \end{array} \right)^T \left( \begin{array}{} \frac{p}{sgn(cC)} \\ \frac{q}{sgn(cF)} \\ \frac{r}{sgn(cI)}) \end{array} \right ) \right| }{{\left| \left ( \begin{array}{} cC \\ cF \\ cI \end{array}  \right )^T \left ( \begin{array}{} \frac{p}{sgn(cC)} \\ \frac{q}{sgn(cF)} \\ \frac{r}{sgn(cI)} \end{array}  \right )  \right| }}$
$yI2=\frac{\left| \left ( \begin{array}{} cB \\ cE \\ cH \end{array} \right)^T \left( \begin{array}{} \frac{p}{sgn(cC)} \\ \frac{q}{sgn(cF)} \\ \frac{r}{sgn(cI)}) \end{array} \right ) \right| }{{\left| \left ( \begin{array}{} cC \\ cF \\ cI \end{array}  \right )^T \left ( \begin{array}{} \frac{p}{sgn(cC)} \\ \frac{q}{sgn(cF)} \\ \frac{r}{sgn(cI)} \end{array}  \right )  \right| }}$
IncenterThreeLinesEquations
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