What is the point of intersection of perpendicular bisector and angle bisector of a triangle called?
Here's the question: In a triangle ABC, perpendicular bisector of side AC is $x-y=4$ and angle bisector of A is $3x+4y+k=0$. Vertex C is the incenter of triangle formed by $(0,0)$, $(3,0)$ and $(0,4)$. Find equation of AB?
The answer is quite a long process of finding the points of intersections. The first is to find the coordinate of the incenter which can be found out from:
$(Ox,Oy)=(\frac{aAx+bBx+cCx}{p},\frac{aAy+bBy+cCy}{p})$
with $p= perimeter$ $;a,b,c=side$ $lengths$ $opposite$ $to$ $A(Ax,Ay),B(Bx,By),C(Cx,Cy)$
$(Ox,Oy)=(1,1)$ for the given problem, which is the coordinate of the vertex $C$.
The equation of the line perpendicular to $x-y=4$ and passes through the point $(1,1)$ is $y+x=2$.
Therefore, the mirror of the point $(1,1)$ through the line $y-x=4$ is $(5,-3)$.
But, $3x+4y+k=0$ passes through the point $(5,-3)$ and solving gives $k=-3$.
As Line $3x+4y-3=0$ is the anglebisector , we can use the mirror concept using this line as the mirror and find out the equation of line $AB$ i.e the line that passes through $(5,-3)$ and the mirror of $(1,1)$ through the line $3x+4y-3=0$ as the mirror, and the mirror point is $(p,q)=(\frac{1}{25},\frac{-7}{25})$.
Therefore, the line through $(5,-3)$ and $(\frac{1}{25},\frac{-7}{25})$ will be the equation of the line $AB$