Find the incentre of the triangle

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 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.

Answer 2

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.

Answer 3

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 $(1,\;1)$

Answer 4

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 $D(1,1)$.

Answer 5

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 $\color{blue}{I\left(1,1\right)}$.