Question: > 
> Find the incentre of the triangle in the $xy-$plane whose sides are given
>  by the lines $x = 0, y = 0$ and $x/3 + y/4 = 1$
> 
> 
> 

I was trying this question many times but i could not get the solution. From my point of view, my incentre coordinate $(0+3+0/3 ,4+0+)/3) =(1,4/3)$

I don't know whether my answer is correct or not.

If anybody could help me, I would be very thankful to him.

Answer: 
It is well know that in triangle with some angle $90^{\circ}$ and sides $a,b,c$ and $r$ as radius of incircle we have $$r = {a+b-c\over 2}$$ if $c$ is hypotenuse. Thus $r = 1$ and so $(1,1)$ is the answer.
The inradius of a right triangle with side lengths $3,4,5$ clearly is $1$, from $(a+b+c)r=2\Delta$.  

Since the incenter has to lie on the $x=y$ line, it is located at $\color{blue}{I\left(1,1\right)}$.