The incentre of the triangle formed by the li | vedclass

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(C) The lines $x=1$ and $y=1$ intersect at $B(1, 1)$. The line $x+y=1$ intersects $x=1$ at $A(1, 0)$ and $y=1$ at $C(0, 1)$.The vertices of the triang

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$\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$

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(C) The lines $x=1$ and $y=1$ intersect at $B(1, 1)$. The line $x+y=1$ intersects $x=1$ at $A(1, 0)$ and $y=1$ at $C(0, 1)$.The vertices of the triangle are $A(1, 0)$,$B(1, 1)$,and $C(0, 1)$.The lengths of the sides are:$a = BC = \sqrt{(1-0)^2 + (1-1)^2} = 1$$b = CA = \sqrt{(1-0)^2 + (0-1)^2} = \sqrt{1+1} = \sqrt{2}$$c = AB = \sqrt{(1-1)^2 + (1-0)^2} = 1$The incentre $(I_x, I_y)$ is given by:$I_x = \frac{ax_1 + bx_2 + cx_3}{a+b+c} = \frac{1(1) + \sqrt{2}(1) + 1(0)}{1+\sqrt{2}+1} = \frac{1+\sqrt{2}}{2+\sqrt{2}} = \frac{1+\sqrt{2}}{\sqrt{2}(\sqrt{2}+1)} = \frac{1}{\sqrt{2}}$$I_y = \frac{ay_1 + by_2 + cy_3}{a+b+c} = \frac{1(0) + \sqrt{2}(1) + 1(1)}{1+\sqrt{2}+1} = \frac{1+\sqrt{2}}{2+\sqrt{2}} = \frac{1}{\sqrt{2}}$Thus,the incentre is $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$.

The incentre of the triangle formed by the lines $x+y=1$,$x=1$,and $y=1$ is

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