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?

Calculation:

Given points A(1,1), B(0,0), and C(2,0). The angle bisectors meet at P (the incenter).

Compute the side lengths:

\(a = BC = \sqrt{(2 - 0)^{2} + (0 - 0)^{2}} = 2\)

\(b = CA = \sqrt{(1 - 2)^{2} + (1 - 0)^{2}} = \sqrt{2}\)

\(c = AB = \sqrt{(1 - 0)^{2} + (1 - 0)^{2}} = \sqrt{2}\)

Therefore, \(a + b + c = 2 + \sqrt{2} + \sqrt{2} = 2 + 2\sqrt{2}.\)

By the incenter formula,

\(X_{P} \;=\; \frac{a\,x_{A} + b\,x_{B} + c\,x_{C}}{a + b + c} \;=\; \frac{\,2\cdot 1 + \sqrt{2}\cdot 0 + \sqrt{2}\cdot 2\,}{\,2 + 2\sqrt{2}\,} \;=\; \frac{2 + 2\sqrt{2}}{2 + 2\sqrt{2}} \;=\; 1.\)

\(Y_{P} \;=\; \frac{a\,y_{A} + b\,y_{B} + c\,y_{C}}{a + b + c} \;=\; \frac{\,2\cdot 1 + \sqrt{2}\cdot 0 + \sqrt{2}\cdot 0\,}{\,2 + 2\sqrt{2}\,} \;=\; \frac{2}{2 + 2\sqrt{2}} \;=\; \frac{1}{\,1 + \sqrt{2}\,} \;=\; \sqrt{2} \;-\; 1. \)

∴ The incenter is \(P = \bigl(1,\;\sqrt{2} - 1\bigr) \).

Hence, the correct answer is Option 1.