Question

The equations of the sides of a triangle $ABC$ are $BC:7x-y+3=0,CA=x+y+1=0$ and $AB:x-y+3=0$.
Find the centre of the circle described to side $BC$.

Answer 1

The internal bisector of angle $A$ is found to be $y-1=0$
The bisectors of $\angle C$ are $3x+4y+2=0$ and $x-3y+1=0$. The points $A$ and $B$ when put in 1st give results of opposite signs and hence it is internal. Therefore $x-3y+1=0$ is external. You may see that points $A$ and $B$ when put give -ive sign. Similarly bisectors of $\angle B$ are $2x-y+3=0$ and $x+2y-6=0$. Points $A$ and $C$ when put in 1st give both opposite signs and hence they are on the opposite sides. Therefore it is internal and as such $x+2y-6=0$ is external. Hence one internal and two external bisectors are $y-1=0,x-3y+1=0$ and $x+2y-6=0$. All these three are concurrent at the point $(2,1)$