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

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Aicharya · Accepted Answer

From the figure, $A\left(x_1, y_1ight)=(1,0)$ (0, 1$)$ $\begin{aligned} B\left(x_2, y_2ight) & =(1,1) \ 	ext { and } C\left(x_3, y_3ight) & =(0,1) \ a=B C & =\sqrt{(1-0)^2+(1-1)^2}=1 \ b=C A & =\sqrt{(1-0)^2+(0-1)^2}=\sqrt{1+1}=\sqrt{2} \ c=A B & =\sqrt{(1-1)^2+(1-0)^2}=1\end{aligned}$ Incentre of triangle $\begin{aligned} & =\left(\frac{a x_1+b x_2+c x_3}{a+b+c}, \frac{a y_1+b y_2+c y_3}{a+b+c}ight) \ & =\left(\frac{1+\sqrt{2}}{1+1+\sqrt{2}}, \frac{\sqrt{2}+1}{1+1+\sqrt{2}}ight)=\left(\frac{1+\sqrt{2}}{2+\sqrt{2}}, \frac{1+\sqrt{2}}{2+\sqrt{2}}ight) \ & =\left(\frac{1+\sqrt{2}}{\sqrt{2}(1+\sqrt{2})}, \frac{1+\sqrt{2}}{\sqrt{2}(1+\sqrt{2})}ight)=\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}ight)\end{aligned}$

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

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