The x - coordinate of the incentre of the triangle that has the...

Let $\left({x}_{1}, {y}_{1}\right), \left({x}_{2}, {y}_{2}\right)$ and \left({x}_{3}, {y}_{3}\right)$be the vertices of a triangle,</p><p>So,<br />${x}_{1} + {x}_{2} = 0, {x}_{2} + {x}_{3} = 2, {x}_{3} + {x}_{1} = 2$[mid point formula$\dfrac{{x}_{1} + {x}_{2}}{2}$]<br /><br />Similarly,</p><p>${y}_{1} + {y}_{2} = 2, {y}_{2} + {y}_{3} = 2, {y}_{3} + {y}_{1} = 0$<br /><br />So, the vertices are \left(0 ,0\right), \left(0 ,2\right), \left(2, 0\right)$.

We know that $x$-coordinate of incentre $\overset{¯}{x} = \dfrac{a{x}_{1} + b{x}_{2} + c{x}_{3}}{a + b + c}$

$a = 2, b = 2, c = 2\sqrt{2}, {x}_{1} = 0, {x}_{2} = 2, {x}_{3} = 0$

$∴ \overset{¯}{x} = \dfrac{0 + 4 + 0}{2 + 2 + 2\sqrt{2}}$

⇒ \overset{¯}{x} = \left(2 - \sqrt{2}\right)