the incenter of triangle formed by lines y=IxI y=1 is

The figure is shown above and now as we know that OA and OB will be equal and solving y =|x| and y=1 we get, 
A as (-1,1) and B as (1,1)  so calculating AB, 
$$\begin{gathered} \mathrm{AB} =1+1=2 \mathrm{unit} \\ \mathrm{OA}=\sqrt{{\left(1-0\right)}^2+{\left(1-0\right)}^2}=\sqrt{2} \\ \mathrm{And} \mathrm{OA}= \mathrm{OB} =\sqrt{2} \\ \mathrm{Now} \mathrm{as} \mathrm{we} \mathrm{know} \mathrm{that} \mathrm{Area} \mathrm{of} \mathrm{triangle} \mathrm{in} \mathrm{terms} \mathrm{of} \mathrm{inradius} \\ =\mathrm{rs} \\ \mathrm{where} \mathrm{s} \mathrm{is} \mathrm{semi} \mathrm{perimeter}, \\ \mathrm{So} \mathrm{s} = \frac{\sqrt{2}+\sqrt{2}+2}{2}=1+\sqrt{2} \\ \mathrm{So} \mathrm{area} \mathrm{of} \mathrm{triangle} = \mathrm{r}\left(1+\sqrt{2}\right)....................\left(1\right) \\ \mathrm{As} \mathrm{here} \mathrm{in} \mathrm{triangle} \mathrm{OM}=1\mathrm{unit} \mathrm{is} \mathrm{perpendicular} \mathrm{and} \mathrm{AB}\hspace{0.17em}\mathrm{is} \mathrm{base} \mathrm{so} , \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} =\frac{1}{2}\times \mathrm{AB}\times \mathrm{OM} \\ =\frac{1}{2}\times 2\times 1=1 {\mathrm{unit}}^2.............................\left(2\right) \\ \mathrm{From} \left(1\right) \mathrm{and} \left(2\right) \mathrm{we} \mathrm{get}, \\ \mathrm{r}\left(1+\sqrt{2}\right)=1 \\ \mathrm{r} = \frac{1}{\left(1+\sqrt{2}\right)}\times \frac{\left(-1+\sqrt{2}\right)}{\left(-1+\sqrt{2}\right)}=\left(\sqrt{2}-1\right) \mathrm{unit} \end{gathered}$$