What is the centroid of a triangle with vertices at #(a, b)#, #(c, d)#, and #(e, f)#?

Question

Mason Adams · Accepted Answer

The centroid is the average of the coordinates:

#({a+c+e}/3, {b+d+f}/3)# The point where the medians intersect is known as the centroid of a triangle. The line segments that join each vertex to the opposite side's midpoint are known as medians. Theorem: A triangle's medians are contemporaneous. If three medians are concurrent, then they meet at some point. Proof: Just solve the median equations and find the points where the pairs intersect. Since parametric forms are simple to write, let's use them. Median 1 endpoints #(a,b)# and #( {c+e}/2, {d+f}/2 ) # is line segment # (x,y) = (1-t)(a,b) + t( {c+e}/2, {d+f}/2 ) quad # for # 0 \le t le 1.# Median 2: #(c,d)# and #( {a+e}/2, {b+f}/2 ) # # (x,y) = (1-u)(c,d) + u( {a+e}/2, {b+f}/2 ) quad # for # 0 \le u le 1.# Median 3: #(e,f)# and #( {a+c}/2, {b+d}/2 ) # # (x,y) = (1-v)(e,f) + v( {a+c}/2, {b+d}/2 ) quad # for # 0 \le v le 1.# When do medials 1 and 2 meet? # (1-t)(a,b) + t( {c+e}/2, {d+f}/2 ) = (1-u)(c,d) + u( {a+e}/2, {b+f}/2 ) # # 2(1-t)(a,b) + t( c+e, d+f ) = 2(1-u)(c,d) + u( a+e, b+f ) # # 2(a,b) + t(c+e-2a,d+f-2b) = 2(c,d) + u(a+e-2c,b+f-2d)# Two equations there, # 2a + t(c+e -2a ) = 2c+ u(a+e - 2c) # # 2b + t(d+f -2b ) = 2d+ u(b+f - 2d) # We solve for #t#, # t(c+e -2a )(b+f - 2d) = (2c-2a) (b+f - 2d)+ u(a+e - 2c)(b+f - 2d) # # t(d+f -2b )(a+e - 2c) = (2d-2b)(a+e - 2c) + u(b+f - 2d)(a+e - 2c) # # t ((c+e -2a )(b+f - 2d) - (d+f -2b )(a+e - 2c) ) = 2(c-a) (b+f - 2d) - 2(d-b)(a+e - 2c) # # t ( ( bc + be - 2ab +cf +ef -2af -2cd -2de +4ad) - (ad + af - 2 ab +de + df -2be -2cd -2cf + 4bc) ) = 2 (bc + cf -2cd - ab -af +2ad -( ad + de -2cd -ab -be +2bc) ) # # t ( - 3 bc +3 be + 3 cf -3af -3de +3ad) = 2 ( - bc + cf -af +ad -de+ be -bc) ) # # 3t ( - bc + be + cf - af - de + ad) = 2 ( - bc + cf -af +ad -de+ be -bc) ) # # t = 2/3 # Wow, Just by symmetry we'll get #u=2/3# and #v=2/3# as well. This indicates that the centroid is # (1-2/3)(a,b) + 2/3( {c+e}/2, {d+f}/2 ) = ( {a+c+e}/3, {b+d+f}/3 ) # Naturally, if we had applied the first and third or second and third median equations, we would have obtained the same result. Not only are the medians concurrent, meeting at a common point called the centroid, but we showed the centroid divides each median in the ratio #2:1#, the longer piece to the vertex.

Theodore Abad · Answer

undefined The centroid of a triangle with vertices at $(a, b)$, $(c, d)$, and $(e, f)$ is given by the coordinates:

$$ \left( \frac{a + c + e}{3}, \frac{b + d + f}{3} \right) $$