The vertices of a triangle are at (2,2), (0,6) and (8,10). Find the coordinates of the trisection point of each median which is nearer the opposite side.

Let (2, 2), (0, 6) and (8, 10) be the vertices A, B and C of the triangle respectively. Let AD, BE, CF be the medians

The coordinates of D are:

D = (4, 8)

The coordinates of E are:

E = (5, 6)

The coordinates of F are:

F = (1, 4)

Let P be the trisection point of the median AD which is nearer to the opposite side BC

∴ P divides DA in the ratio 1:2 internally

Let Q be the trisection point of the median BE which is nearer to the opposite side CA

∴ Q divides EB in the ratio 1:2 internally

Let R be the trisection point of the median CF which is nearer to the opposite side AB

∴ R divides FC in the ratio 1:2 internally

Therefore, Coordinates of required trisection points are