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incentre of the triangle the coordinates of whose vertices are given by A(x1, y1), B(x2, y2), C(x3, y3).
By geometry, we know that BD/DC = AB/AC (since AD bisects ÐA).
If the lengths of the sides AB, BC and AC are c, a and b respectively, then BD/DC = AB/AC = c/b.
Coordinates of D are (bx2+cx3/b+c, by2+cy3/b+c)
IB bisects DB. Hence ID/IA = BD/BA = (ac/b+c)/c = a/c+b.
Let the coordinates of I be (x, y).
Then x = ax1+bx2+cx3/a+b+c, y = ay1+by2+cy3/a+b+c.
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