
To find the centroid of a triangle whose vertices are given Let A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) be the vertices of the triangle ABC. Let AD be the median bisecting the base. Then D = {(x_{2} + x_{3})/2, (y_{2} + y_{3})/2} Now the point G on AD, which divides it internally in the ratio 2:1, is the centroid. If (x,y) are the coordinates of G, then x = 2x(x_{2}+x_{3})/2 + (1 x x_{1})/2+1 = (x_{1}+x_{2}+x_{3})/3 y = 2x(y_{2}+y_{3})/2 + (1 x y_{1})/(2+1) = (y_{1}+y2+y3)/3 Hence, the coordinates of the centroid are given by x = (x_{1}+x_{2}+x_{3})/3, y = (y_{1}+y_{2}+y_{3})/3. Directions: Solve the following. 