The correct option is D (1,1√3)
Let the vertices be A(1,√3),B(0,0),C(2,0)
Now,
AB=√(1−0)2+(√3−0)2=2 units
BC=√(0−2)2+(0−0)2=2 units
CA=√(2−1)2+(0−√3)2=2 units
Now, incentre
=(ax1+bx2+cx3a+b+c,ay1+by2+cy3a+b+c)=(x1+x2+x33,y1+y2+y33)(∵a=b=c)=(1+0+23,√3+0+03)
=(1,1√3)