Question

Find the coordinates of the incentre of the triangle, equations of whose sides are
$x+1=0,-3x+4y+5=0,5x+12y=27$

Answer 1

By solving the equations of given lines find the coordinates of the vertices as $A(3,1),B(-1,-2),C(-1,8/3)$
The bisectors of $\angle A$ are given by
$\frac{3x-4y-5}{5}=\pm \frac{5x+12y-27}{13}$
or $x-8y+5=0$ and $8x+y-25=0$
Since we have to find the internal bisector of angle $A$, the points $B(-1,-2)$ and $C(-1,8/3)$ must lie on the opposite sides of the internal bisector. Putting in any equation of the bisector say in $x-8y+5=0$, we get $20,-\frac{52}{3}$ i.e of opposit signs and hence this bisector is internal bisector of angle $A$. It may be verified that $B$ and $C$ when put in 2nd bisector will give results of the same sign. Similarly the internal bisector of angle $C$ is found to be $9x+6y-7=0$. Solving these two internal bisectors, we get the coordinates of incentre as $(\frac{1}{3},\frac{2}{3})$