The x - coordinate of the incentre of the triangle that has the coordinates of midpoints of its sides as…

Let $\left(x_1, y_1\right), \left(x_2, y_2\right)$ and $\left(x_3, y_3\right)$ be the vertices of a triangle,

So,
$x_1+x_2=0, x_2+x_3=2, x_3+x_1=2$ [mid point formula $\frac{x_1+x_2}{2}$ ]

Similarly,

$y_1+y_2=2, y_2+y_3=2, y_3+y_1=0$

So, the vertices are $\left(0,0\right), \left(0,2\right), \left(2,0\right)$.

We know that $x$-coordinate of incentre $\overline{x}=\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c}$   

$a=2,b=2,c=2\sqrt{2},x_1=0,x_2=2,x_3=0$

$\therefore \overline{x}=\frac{0+4+0}{2+2+2\sqrt{2}}$

$\Rightarrow \overline{x}=\left(2-\sqrt{2}\right)$