Question

Let the sides of $△ABC$ be $3x+4y=0,4x+3y=0$ and $x=3$. If $(h,k)$ be the centre of the circle inscribed in $△ABC$, then the value of $(h+k)$ is

Answer 1

Equations of angle bisectors of $\angle A$, we get
$\frac{3x+4y}{\sqrt{3^2+4^2}}=\pm \frac{4x+3y}{\sqrt{3^2+4^2}}\Rightarrow 3x+4y=\pm (4x+3y)\Rightarrow x=\pm y$
By observation, we see that the internal angle bisector of $A$ has negative slope, so the equation of internal bisector is
$x=-y\cdots \left(1\right)$
Equation of obtuse angle bisector of $\angle B$, we get
$\frac{3x+4y}{\sqrt{3^2+4^2}}=\pm \frac{x-3}{1}\Rightarrow 3x+4y=\pm 5(x-3)\Rightarrow 3x+4y=\pm (5x-15)\Rightarrow 3x+4y=5x-15\text{or}3x+4y=-5x+15\Rightarrow 2x-4y-15=0\text{or}8x+4y-15=0$
By observation, we see that the external angle bisector of $A$ has positive slope, so the equation of external bisector is
$2x-4y-15=0\cdots \left(2\right)$
Solving equation $\left(1\right)$ and $\left(2\right),$ we get the incentre
$-2y-4y-15=0\Rightarrow 6y=-15\Rightarrow y=-\frac{5}{2}\Rightarrow x=\frac{5}{2}$
So, the incentre $(h,k)=(\frac{5}{2},-\frac{5}{2})$
$\therefore h+k=0$