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: 
Let $A(3,0)$, $B(0,4)$ and $O(0,0)$.

Now, let the incircle touches to $OA$, $OB$ and $AB$ in the points $K$, $M$ and $N$ respectively. 

Also, let $r$ be a radius of the incircle and $D$ be a center of the the incircle. 

Thus, easy to see that $DMOK$ is square with side-length $r$ and $AK=AN$, $BM=BN$.

Now, since $AB=\sqrt{3^2+4^2}=5$ and $AN+BN=AB$, we obtain:
$$4-r+3-r=5$$ or 
$$r=1,$$
which gives $D(1,1)$.
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.