Blog: 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.

$\displaystyle{d}=\sqrt{{{{\left({x}_{{2}}-{x}_{{1}}\right)}}^{{2}}+{{\left({y}_{{2}}-{y}_{{1}}\right)}}^{{2}}}}$

For side AB, its length is:

$\displaystyle{d}_{{{A}{B}}}=\sqrt{{{{\left({0}-{1}\right)}}^{{2}}+{{\left({1}-{1}\right)}}^{{2}}}}$

$\displaystyle{d}_{{{A}{B}}}={1}$

For side BC, its length is:

$\displaystyle{d}_{{{B}{C}}}=\sqrt{{{{\left({1}-{1}\right)}}^{{2}}+{{\left({1}-{0}\right)}}^{{2}}}}$

$\displaystyle{d}_{{{B}{C}}}={1}$

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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.

$\displaystyle{d}=\sqrt{{{{\left({x}_{{2}}-{x}_{{1}}\right)}}^{{2}}+{{\left({y}_{{2}}-{y}_{{1}}\right)}}^{{2}}}}$

For side AB, its length is:

$\displaystyle{d}_{{{A}{B}}}=\sqrt{{{{\left({0}-{1}\right)}}^{{2}}+{{\left({1}-{1}\right)}}^{{2}}}}$

$\displaystyle{d}_{{{A}{B}}}={1}$

For side BC, its length is:

$\displaystyle{d}_{{{B}{C}}}=\sqrt{{{{\left({1}-{1}\right)}}^{{2}}+{{\left({1}-{0}\right)}}^{{2}}}}$

$\displaystyle{d}_{{{B}{C}}}={1}$

And for side AC, its length is:

$\displaystyle{d}_{{{A}{C}}}=\sqrt{{{{\left({1}-{0}\right)}}^{{2}}+{{\left({0}-{1}\right)}}^{{2}}}}{\)}$

$\displaystyle{d}_{{{A}{C}}}=\sqrt{{2}}$

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)

$\displaystyle{h}=\frac{{{a}{A}_{{x}}+{b}{B}_{{x}}+{c}{C}_{{x}}}}{{{a}+{b}+{c}}}$

$\displaystyle{k}=\frac{{{a}{A}_{{y}}+{b}{B}_{{y}}+{c}{C}_{{y}}}}{{{a}+{b}+{c}}}$

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:

$\displaystyle{h}=\frac{{{1}\cdot{0}+\sqrt{{2}}\cdot{1}+{1}\cdot{1}}}{{{1}+\sqrt{{2}}+{1}}}=\frac{{\sqrt{{2}}+{1}}}{{\sqrt{{2}}+{2}}}$

$\displaystyle{h}=\frac{{\sqrt{{2}}+{1}}}{{\sqrt{{2}}+{2}}}\cdot\frac{{\sqrt{{2}}-{2}}}{{\sqrt{{2}}-{2}}}=\frac{{{2}-{2}\sqrt{{2}}+\sqrt{{2}}-{2}}}{{{2}-{2}\sqrt{{2}}+{2}\sqrt{{2}}-{4}}}=\frac{{-\sqrt{{2}}}}{{-{2}}}$