# What is the incenter of the triangle with vertices (1, 4), (3, 8) and (4, 2).

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### 1 Answer

For a circle with vertices (xa, ya), (xb, yb) and (xc, yc) the coordinates of the in-center is given by `((a*xa + b*xb + c*xc)/P, (ya + yb + yc)/P)` where a, b and c are the the length of the side opposite the points (xa, ya), (xb, yb) and (xc, yc) respectively and P = a + b + c.

The vertices of the triangle are (1, 4), (3, 8) and (4, 2).

The length of the side opposite (1, 4) is `sqrt((4-3)^2 + (8 - 2)^2) = sqrt 37`

The length of the side opposite (3, 8) is `sqrt((4 - 1)^2 + (4 - 2)^2) = sqrt 13`

The length of the side opposite (4, 2) is `sqrt((3 - 1)^2 + (8 - 4)^2) = sqrt 20`

Substituting these values in the expression for the in-center.

`((1*sqrt 37 + 3*sqrt 13 + 4*sqrt 20)/(sqrt 37 + sqrt 13 + sqrt 20), (4*sqrt 37 + 8*sqrt 13 + 2*sqrt 20)/(sqrt 37 + sqrt 13 + sqrt 20))`

=> (2.4566, 4.3868)

**The required in-center is (2.4566, 4.3868)**

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