Question 1041673
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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 side.
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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 <A HREF=https://www.algebra.com/algebra/homework/word/geometry/The-Centroid-of-a-triangle-is-the-Intersection-point-of-its-medians.lesson>The Centroid of a triangle is the Intersection point of its medians</A> in this site.

And its coordinates are 

{{{x[0]}}} = {{{(x[A] + x[B] + x[C])/3}}} = {{{(4+10+2)/3}}} = {{{16/3}}} = {{{5}}}{{{1/3}}},

{{{y[0]}}} = {{{(y[A] + y[B] + y[C])/3}}} = {{{(-4-4+6)/3}}} = {{{-4/3}}} = {{{-1}}}{{{1/3}}}.
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