The vertices of a triangle are $A(1, 1)$, $B(0, 0)$ and $C(2, 0)$. The angular bisectors of the triangle meet at $P$. What are the coordinates of $P$?

The point of intersection of the angular bisectors of a triangle is its incenter.

The vertices of the triangle are $A(1, 1)$, $B(0, 0)$, and $C(2, 0)$.

The lengths of the sides of the triangle are:
$a = BC = \sqrt{(2-0)^2 + (0-0)^2} = 2$
$b = AC = \sqrt{(2-1)^2 + (0-1)^2} = \sqrt{2}$
$c = AB = \sqrt{(1-0)^2 + (1-0)^2} = \sqrt{2}$

The coordinates of the incenter $P(x, y)$ are given by:
$x = \dfrac{a x_A + b x_B + c x_C}{a + b + c}$
$y = \dfrac{a y_A + b y_B + c y_C}{a + b + c}$

Substituting the values:
$x = \dfrac{2(1) + \sqrt{2}(0) + \sqrt{2}(2)}{2 + \sqrt{2} + \sqrt{2}} = \dfrac{2 + 2\sqrt{2}}{2 + 2\sqrt{2}} = 1$

$y = \dfrac{2(1) + \sqrt{2}(0) + \sqrt{2}(0)}{2 + \sqrt{2} + \sqrt{2}} = \dfrac{2}{2 + 2\sqrt{2}} = \dfrac{1}{\sqrt{2} + 1}$

Rationalizing the denominator for $y$:
$y = \dfrac{1}{\sqrt{2} + 1} \times \dfrac{\sqrt{2} - 1}{\sqrt{2} - 1} = \sqrt{2} - 1$

The coordinates of $P$ are $(1, \sqrt{2} - 1)$.

Answer: $(1, \sqrt{2}-1)$