Centroidof TriangleThe centroid of a triangle is the point of concurrency of the medians. The centroid G of the triangle ABC, divides the median AD, in the ratio of 2 : 1.Illustration:Find the centroid of the triangle the coordinates of whose vertices are given by A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) respectively.Solution:AG/AD = 2/1Since D is the midpoint of BC, coordinates of D are (x_{2}+x_{3}/2, y_{2}+y_{3}/2)Using the section formula, the coordinates of G are(2(x_{2}+x_{3}/2)+1.x_{1}/2+1, 2(y_{2}+y_{3}/2)+1.y_{1}/2+1)⇒ Coordinates of G are (x1+x2+x3/3, y1+y2+y3/3).Incentreof TriangleThe incentre ‘I’ of a triangle is the point of concurrency of the bisectors of the angles of the triangle.Illustration:Find the incentre of the triangle the coordinates of whose vertices are given by A(x_{1}, y_{1}), B(x_{2}, y_{2}), C(x_{3}, y_{3}).Solution: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 (bx_{2}+cx_{3}/b+c, by_{2}+cy_{3}/b+c)IB bisects ÐB. Hence ID/IA = BD/BA = (ac/b+c)/c = a/c+b._{ – –}Let the coordinates of I be (x, y)._{– }_{ –}Then x = ax_{1}+bx_{2}+cx_{3}/a+b+c, y = ay_{1}+by_{2}+cy_{3}/a+b+c.

Circum Centreof TriangleThis the point of concurrency of the perpendicular bisectors of the sides of the triangle. This is also the centre of the circle, passing through the vertices of the given triangle.

Orthocentreof TriangleThis is the point of concurrency of the altitudes of the triangle.ExcentreExcentre of a triangle is the point of concurrency of bisectors of two exterior and third interior angle. Hence there are three excentres I_{1}, I_{2}and I_{3}opposite to three vertices of a triangle.If A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) are the vertices of a triangle ABC,coordinates of centre of ex-circle opposite to vertex A are given asI_{1}(x, y) = (–ax_{1}+bx_{2}+cx_{3}/a+b+c/–a+b+c, –ay_{1}+by_{2}+cy_{3}/–a+b+c).Similarly co-ordinates of centre of I_{2}(x, y) and I_{3}(x, y) areI_{2}(x, y) = (ax_{1}–bx_{2}+cx_{3}/a–b+c, ay_{1}–by_{2}+cy_{3}/a–b+c)I

_{3}(x, y) = (ax_{1}+bx_{2}–cx_{3}/a+b–c, ay_{1}+by_{2}–cy_{3}/a+b–c)