Find the incenter of a triangle formed by x+y=1,x=1,y=1

First, draw the triangle formed by the three equations x+y=1, x=1 and y=1.

Let the vertices of the triangle be A, B and C (see attached figure).
Base on the graph, the coordinates of the vertices are:
A(0,1)
B(1,1) and
C(1,0)
To determine the length of each side of the triangles, apply the distance formula.
d=sqrt((x_2 - x_1)^2+ (y_2-y_1)^2)
For side AB, its length is:
d_(AB)=sqrt((0-1)^2+(1-1)^2)
d_(AB)=1
For side BC, its length is:
d_(BC)=sqrt((1-1)^2+(1-0)^2)
d_(BC)=1
And for side AC, its length is:
d_(AC)=sqrt((1-0)^2+(0-1)^2))
d_(AC)=sqrt2
Now that the coordinates of the vertices and the length of each sides are known, apply the formula below to solve for the coordinates of the incenter (h,k)
h= (aA_x+bB_x+cC_x)/(a+b+c)
k=(aA_y+bB_y+cC_y)/(a+b+c)
where
Ax & Ay are coordinates of vertex A,
Bx & By are coordinates of vertex B,
Cx & Cy are coordinates of vertex C,
a is the length of the side opposite vertex A (which is side BC),
b is the length of the side opposite vertex B (which is side AC), and
c is the length of the side opposite vertex C (which is side AB).
So the values of h and k are:
h = (1*0 + sqrt2*1+1*1)/(1+sqrt2+1)=(sqrt2+1)/(sqrt2+2)
h=(sqrt2+1)/(sqrt2+2)*(sqrt2-2)/(sqrt 2-2)= (2-2sqrt2+sqrt2-2)/(2-2sqrt2+2sqrt2-4) = (-sqrt2)/(-2)
h=sqrt2/2
 
k=(1*1+sqrt2*1+1*0)/(1+sqrt2+1) =(sqrt2+1)/(sqrt2+2)
k=sqrt2/2
 
Therefore, the incenter of the triangle is (sqrt2/2,sqrt2/2) .