The in-centre of the triangle with vertices $(0, 0), (1, 0), (0, 1)$ is

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(\frac{2 - \sqrt{2}}{2}, \frac{2 - \sqrt{2}}{2})

(\frac{\sqrt{2} - 1}{2}, \frac{\sqrt{2} - 1}{2})

(\frac{1}{3}, \frac{1}{3})

(\frac{2 + \sqrt{2}}{2}, \frac{2 + \sqrt{2}}{2})

answer is B.

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Detailed Solution

Given vertices

A (0, 0), B (1, 0), C (0, 1)

$x_{1} y_{1} x_{2} y_{2} x_{3} y_{3}$

Incentre $(I) = [\frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c}, \frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c}]$

$a = B C = \sqrt{1 + 1} = \sqrt{2}$

$b = A C = \sqrt{1} = 1$

$c = A B = \sqrt{1} = 1$

$I = [\frac{(\sqrt{2}) (0) + (1) (1) + 1 (0)}{2 + \sqrt{2}}, \frac{\sqrt{2} (0) + 1 (0) + 1 (1)}{2 + \sqrt{2}}]$

$= [\frac{1}{2 + \sqrt{2}}, \frac{1}{2 + \sqrt{2}}]$

Now rationalize the coordinates

$= [\frac{2 - \sqrt{2}}{4 - 2}, \frac{2 - \sqrt{2}}{4 - 2}]$

$= [\frac{2 - \sqrt{2}}{2}, \frac{2 - \sqrt{2}}{2}]$

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