Question

The incentre of the triangle formed by the lines $y=\mid x\mid$ and $y=1$ is:

• A. $(0,2-\sqrt{2})$
• B. $(2-\sqrt{2},0)$
• C. $(2+\sqrt{2},0)$
• D. $(0,2+\sqrt{2})$

Answer 1

The correct option is A $(0,2-\sqrt{2})$

Given a triangle with vertices $(x_a,y_a),(x_b,y_b),(x_c,y_c)$ and the sides opposite those vertices as $a,b,c$

Then, the coordinates for the incenter are given by:

$\frac{(a{x}_a+b{x}_b+c{x}_c)}{p},\frac{(a{y}_a+b{y}_b+c{y}_c)}{p}$ where $p$ is the perimeter of the triangle.

The triangle given is formed by $y=\left|x\right|$ and $y=1$. The verices are $A=(-1,1),B=(1,1)$ and $C=(0,0)$ with sides $a=b=\sqrt{2},c=2$. The perimeter is $2+2\sqrt{2}$.

The coordinates of the incenter are:

$(\frac{\sqrt{2}(-1)+\sqrt{2}\left(1\right)+2\left(0\right)}{(2+\sqrt{2})},\frac{\sqrt{2}\left(1\right)+\sqrt{2}\left(1\right)+2\left(0\right)}{(2+\sqrt{2})})$

$=(0,\frac{2\sqrt{2}}{(2+\sqrt{2})})=(0,2-\sqrt{2})$ so the answer is A