The incentre of the triangle formed by the li | vedclass

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(A) The lines are $x+y=1$,$x=1$,and $y=1$.Finding the vertices of the triangle:$1$. Intersection of $x=1$ and $y=1$ is $P(1, 1)$.$2$. Intersection of

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

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(A) The lines are $x+y=1$,$x=1$,and $y=1$.Finding the vertices of the triangle:$1$. Intersection of $x=1$ and $y=1$ is $P(1, 1)$.$2$. Intersection of $x+y=1$ and $x=1$ is $A(1, 0)$.$3$. Intersection of $x+y=1$ and $y=1$ is $B(0, 1)$.The side lengths are:$a = BP = \sqrt{(1-0)^2 + (1-1)^2} = 1$$b = AP = \sqrt{(1-1)^2 + (1-0)^2} = 1$$c = AB = \sqrt{(1-0)^2 + (0-1)^2} = \sqrt{2}$The incentre $(I_x, I_y)$ is given by $\left(\frac{ax_1+bx_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c}\right)$:$I_x = \frac{1(1) + 1(1) + \sqrt{2}(1)}{1+1+\sqrt{2}} = \frac{2+\sqrt{2}}{2+\sqrt{2}} = 1$ (Wait,re-evaluating vertices: $P(1,1), A(1,0), B(0,1)$).$I_x = \frac{1(1) + 1(1) + \sqrt{2}(0)}{1+1+\sqrt{2}} = \frac{2}{2+\sqrt{2}} = \frac{2(2-\sqrt{2})}{4-2} = 2-\sqrt{2}$.$I_y = \frac{1(1) + 1(0) + \sqrt{2}(1)}{1+1+\sqrt{2}} = \frac{1+\sqrt{2}}{2+\sqrt{2}} = \frac{1}{\sqrt{2}}$.Actually,the vertices are $P(1,1), A(1,0), B(0,1)$. The incentre is $\left(1-\frac{1}{\sqrt{2}}, 1-\frac{1}{\sqrt{2}}\right)$.

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

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