If the incentre of the triangle formed by the | vedclass

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(C) The lines are $L_1: x-2=0$,$L_2: x+y-1=0$,and $L_3: x-y+3=0$.Intersection of $L_1$ and $L_2$: $x=2, 2+y-1=0 \implies y=-1$. Vertex $A = (2, -1)$.I

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$\frac{2\sqrt{2}-1}{\sqrt{2}+1}$

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(C) The lines are $L_1: x-2=0$,$L_2: x+y-1=0$,and $L_3: x-y+3=0$.Intersection of $L_1$ and $L_2$: $x=2, 2+y-1=0 \implies y=-1$. Vertex $A = (2, -1)$.Intersection of $L_1$ and $L_3$: $x=2, 2-y+3=0 \implies y=5$. Vertex $B = (2, 5)$.Intersection of $L_2$ and $L_3$: $x+y=1$ and $x-y=-3$. Adding gives $2x=-2 \implies x=-1$. Then $-1+y=1 \implies y=2$. Vertex $C = (-1, 2)$.Side lengths: $c = AB = \sqrt{(2-2)^2 + (5-(-1))^2} = 6$.$b = AC = \sqrt{(2-(-1))^2 + (-1-2)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2}$.$a = BC = \sqrt{(2-(-1))^2 + (5-2)^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$.The incentre $(\alpha, \beta)$ is given by $(\frac{ax_1+bx_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c})$.$\beta = \frac{3\sqrt{2}(-1) + 3\sqrt{2}(5) + 6(-1)}{3\sqrt{2} + 3\sqrt{2} + 6} = \frac{-3\sqrt{2} + 15\sqrt{2} - 6}{6\sqrt{2} + 6} = \frac{12\sqrt{2}-6}{6(\sqrt{2}+1)} = \frac{2\sqrt{2}-1}{\sqrt{2}+1}$.

If the incentre of the triangle formed by the lines $x-2=0$,$x+y-1=0$,and $x-y+3=0$ is $(\alpha, \beta)$,then $\beta=$

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