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: 
Hint: with the vertices at $(0,0), (0,4), (3,0)$ the incentre must lie on the bisector of the angle at the origin i.e. $x=y$.

The inradius $r$ satisfies area of triangle $=rs$ where $s$ is half the sum of the sides. To see this connect the incentre to the three vertices, and drop perpendiculars to each of the sides. The lines to the vertices split the figure into three triangles each of height $r$ and base one of the sides of the triangle.

This is enough data to make solving quite easy.
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 $\color{blue}{I\left(1,1\right)}$.