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

Solving the lines    x+y=1, x = 1,  y = 1 we get A (1,1),B (1,0),C (0,1)AB=1=c,     BC=2 =a,     AC=1=b                I=ax1+bx2+cx3a+b+c,ay1+by2+cy3a+b+c  = 2+12(2+1),2+12(2+1)                =(12,12)