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`

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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)` .