The coordinates of the centroid of a triangle are the arithmetic mean of the x and y coordinates of the vertices. How do you prove this?

2 Answers
Somebody N.
Jun 14, 2018

See below.

Explanation:

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The centroid #G# lies #2/3# along the median from each vertex #A, B, C#

Looking at the median #AD#

The co-ordinates of #D# can be found using the midpoint formula:

#D=((x_2+x_3)/2,(y_2+y_3)/2)#

The point #G# divides #AD# in the ratio #2:1#

The co-ordinates of #G# are calculated using the section formula:

The Section Formula

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The point #P# lies on the line #AB#, such that:

#AP:PB=m:n#

#P " is " m/(m+n)xxAB# away from #A#

For #x#:

#x=x_1+m/(m+n)xx(x_2-x_1)#

# \ \ \ =x_1+(mx_2)/(m+n)-(mx_1)/(m+n)#

# \ \ \ \=((m+n)x_1+mx_2-mx_1)/(m+n)#

# \ \ \ \=(mx_2+nx_1)/(m+n)#

For #y#:

#y=y_1+m/(m+n)xx(y_2-y_1)#

# \ \ \ =y_1+(my_2)/(m+n)-(my_1)/(m+n)#

# \ \ \ \=((m+n)y_1+my_2-my_1)/(m+n)#

# \ \ \ \=(my_2+ny_1)/(m+n)#

#:.#

#P=((mx_2+nx_1)/(m+n),(my_2+ny_1)/(m+n))#

Now back to the beginning:

Our ratio is #m:n=2:1#

We are using median #AD#

Using section formula for #bbx# co-ordinate.

#x=(2((x_2+x_3)/2)+x_1)/(2+1)color(white)(88)# ( note# (x_2+x_3)/2# is point D )

#x=(x_1+x_2+x_3)/3#

Using section formula for #bby# co-ordinate.

#y=(2((y_2+y_3)/2)+y_1)/(2+1)#

#y=(y_1+y_2+y_3)/3#

So co-ordinates of the centroid are:

#G=((x_1+x_2+x_3)/3,(y_1+y_2+y_3)/3)#

Note:

To find this we used the fact that the centroid divided the median in the ratio #2:1#. The proof of this wasn't included, but this can be verified by the use of vectors and the fact that the six triangles formed by the medians inside the triangle #ABC# all have the same area.

This is from the mentioned text book. As you can see it isn't a rigorous proof like the one in the other answer, but maybe it will be of some help to you.

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Dean R.
Jun 15, 2018

Triangle ACE, Vertices #A(a,b), C(c,d), E(e,f)#

Midpoints of the sides

#B={A+C}/2=({a+c}/2, {b+d}/2)#

#D={A+E}/2=({a+e}/2, {b+f}/2)#

#F={C+E}/2=({c+e}/2, {d+f}/2)#

We'll do the medians parametrically so we can show the centroid divides the median into segments ratio #2:1#.

Parametric equations for the medians

From #E# to #B={A+C}/2#

# EB= (1-t)E+tB = E + t({A+C}/2 - E) = E + t/2(A+C-2E)#

From #A# to #D={C+E}/2#

# AD= A + u({C+E}/2 - A) = A + u/2(C+E-2A)#

These meet at #G# which we want to show is the centroid:

#G = E + t/2(A+C-2E) = A + u/2(C+E-2A)#

# e + t/2(a+c-2e) = a +u/2(c+e-2a)#

#f + t/2(b+d-2f) = b+ u/2(d+f-2b)#

# t(d+f-2b)(a+c-2e) = 2(d+f-2b)(a-e)+u(d+f-2b)(c+e-2a)#

# t(c+e-2a)(b+d-2f) = 2(c+e-2a)(b-f)+ u(c+e-2a)(d+f-2b)#

#t ( (d+f-2b)(a+c-2e)- (c+e-2a)(b+d-2f)) = 2(d+f-2b)(a-e) - 2(c+e-2a)(b-f) #

Getting messy.

#t = ( 2(d+f-2b)(a-e) - 2(c+e-2a)(b-f) ) / ( (d+f-2b)(a+c-2e)- (c+e-2a)(b+d-2f))#

#t = { 2 ( ( ad + af -2ab -de -ef +2be)-(bc+be-2ab-cf-ef+2af ) ) }/{ ( ad+af-2ab +cd +cf-2bc -2de -2ef +4be ) - ( bc+be-2ab +cd+de-2ad -2fc-2ef+4af ) } #

#t = { 2 ( ad -de + be-bc+cf -af ) }/{ ( 3ad-3af +3cf-3bc -3de +3be ) } #

#t = 2/3 #

Recall #G = E + t(B-E) = t/2(A+C-2E)#

#t=2/3# shows #G# divides the median #EB# into a 2:1 ratio, #EG=2BG.#

#G = E + t/2(A+C-2E) = E+((2/3)/2)(A+C-2E)= E+1/3(A+C-2E)=1/3(A+C+E)#

That shows #G# is the average of the respective coordinates.

Since we'll get the same #2/3# starting from the other vertices, we'll get the same point #G#, i.e. we've shown the medians are concurrent, they meet at a single point, so we're justified in calling #G# the centroid.

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