For the given triangle ABC. Let AD, BE and CF are the medians

We know that, median divides the line segment into two equal parts, so D is the midpoint of BC, therefore,

Coordinates of point D = \( \dfrac{0+6}{2}, \dfrac{4+0}{2}, \dfrac{0+0}{2}\)

Coordinates of point D = (3,2,0)

AD =\( \sqrt{0– 3)^2 + (0 – 2)^2 + (6– 0)^2} \)

AD = \( \sqrt{9+4+36}\)

AD = \( \sqrt{49}\)

AD = 7

Similarly, E is the midpoint of AC,

Coordinates of point E = \( \dfrac{0+6}{2}, \dfrac{0+0}{2}, \dfrac{0+6}{2}\)

Coordinates of point E = (3,0,3)

AC = \( \sqrt{3– 0)^2 + (0 – 4)^2 + (3– 0)^2} \)

AC = \( \sqrt{9+16+9} \)

AC = \( \sqrt{34}\)

Similarly, F is the midpoint of AB,

Coordinates of point F =\( \dfrac{0+0}{2}, \dfrac{0+4}{2}, \dfrac{6+0}{2}\)

Coordinates of point F = (0,2,3)

CF =\( \sqrt{6– 0)^2 + (0 – 2)^2 + (0– 3)^2} \)

CF = \( \sqrt{36+4+9} \)

CF = \( \sqrt{49}\)

CF = 7

Therefore, the lengths of the medians of the triangle ABC we obtain are,

7, \( \sqrt{34}\), 7

A(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle ∠BAC meets BC.

The mid-points of the sides of a triangle ABC are given by (-2, 3, 5), (4, -1, 7) and (6, 5, 3). Find the coordinates of A, B and C.

If the points A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6) are collinear, find the ratio in which C divided AB.

Find the ratio in which the line segment joining the points (2, -1, 3) and (-1, 2, 1) is divided by the plane x + y + z = 5.

Find the ratio in which the line joining (2, 4, 5) and (3, 5, 4) is divided by the yz-plane.