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 $D(1,1)$.
The bisector of the right angle is $y=x$ A point of this bisector has coordinates $P(t;\;t)$ its distance from the angle sides is $|t|$ 

This distance must be equal to the distance from the line $x/3+y/4=1$

The formula of the distance $d$ from a point $(x\_P;\;y\_P)$ to a line $ax+by+c=0$ is
$$d=\dfrac{\left|ax\_p+by\_p+c\right|}{\sqrt{a^2+b^2}}$$

The equation of the line must be written as $4x+3y-12=0$ 

The point has coordinates $(t;\;t)$ and the distance must be $d=|t|$

$\dfrac{\left|3t +4t-12\right|}{\sqrt{3^3+4^2}}=|t|$

$\left|7t-12\right|=5|t|$

$7t-12=5t$ or $7t-12=-5t$

$t=6$ or $t=1$

The result is then $t=1$ and the incentre is $(1,\;1)$