SOLUTION: Given a triangle whose vertices are A(4,-4) B(10,-4) and C(2,6). Find the point on each median that is two-thirds of the distance from the vertex to the midpoint of the opposite si

A median of a triangle IS the line from the vertex to the 
midpoint of the opposite side. 

The 3 medians of every triangle intersect in a point that IS 
two-thirds of the distance from the vertex to the midpoint 
of the opposite side.

Therefore you do not have to be concerned at all with the 
"two-thirds" part.  Where they intersect will automatically 
take care of that fact.  Here's what you do:

1.  Use the midpoint point formula to find the midpoint of
any two of the three sides. [All three medians are drawn 
above in green, but you need only pick two of them]

2.  Find the equations of those two medians using

A. The slope formula.
B. The point slope formula.

3. Solve the system of the two equations of medians by either

A. Substitution, or
B. Elimination. 

4.  The x and y values you get for 3 will be the coordinates
of the desired point.

If you have any trouble, tell me in the thank-you note form 
below and I'll get back to you by email.  No charge ever!
I do this only as a hobby.

Edwin

Actually, this point (intersection of medians of a triangle) is the "centroid of the triangle", or "center of mass of the triangle". 

See the lesson The Centroid of a triangle is the Intersection point of its medians in this site.

And its coordinates are 

 =  =  =  = ,

 =  =  =  = .