Finding the centroid of a triangle
Problem: Prove, using a
vector method, that the medians of a triangle meet at a point which
divides each median in the same ratio. Find this ratio.
This point is called the centroid of the triangle.
A median is a line which joins a vertex of a triangle to the midpoint of
the opposite side.
We have to prove that the medians AP, BQ and CR of triangle ABC do meet
in a single point G and that AG/GP = BG/GQ = CG/GR = k and find the value of
the constant, k.
Suppose that the position vectors of A, B and C with respect to some
origin O are a, b and c.
The right-hand drawing gives the vectors for the sides of the triangle,
found using vector addition.
O need not necessarily lie
in the plane of the triangle.
quite often think
that a, b and c give the sides of the triangle. They
don't! See the drawing below.
For example, AB = -OA + OB.
We start by finding the position vector of the point where 2 of the
medians cut each other.
The median AP lies in the extended line I've shown in my drawing.
(See lines if necessary.)
We'll then have to show that the third median also
passes through this point.
I've chosen to work with AP and BQ.
Suppose that AP and BQ meet at G. We'll need to show that the position
vector of G depends equally on a, b and c. Then, by
similar working, G would also lie on CR proving that the 3 medians do all
meet in a point.
Its vector equation can
be written r = a + sAP.
But AP = AB + BP = -a + b +
1/2(-b + c) = -a + 1/2(b + c).
So the equation of
line AP is r = a + s (-a + 1/2(b + c)).
Show for yourself that the vector equation of line BQ is
r = b + t (-b + 1/2(c + a)).
The 2 lines meet
where r = a + s (-a + 1/2(b + c)) =
r = b + t (-b + 1/2(c + a)).
To get the same vector r for each line, we must have the same
quantity of each of a, b and c.
for c gives 1/2sc = 1/2tc so s = t.
Matching up for a gives
a - sa = 1/2ta so 1 - s = 1/2t giving
t = 2/3 = s.
Therefore G lies 2/3 of the way down both AP and BQ so k = 2/3.
Also, putting s = 2/3 in the equation of line AP gives
r = g = a + 2/3(-a + 1/2(b + c
) = 1/3(a + b + c)).
We see that the position vector r of G depends equally
on a, b and c so therefore, by similar working, G
also lies 2/3 of the way down CR.
The point G is called the centroid
of triangle ABC.
Its position vector g = 1/3 (a + b + c) where
a, b and c are the position vectors of A, B and C.
Physically, if the triangle is made of a uniform thin heavy material, then
G is where the whole weight of the triangle acts.
If the triangle is cut into tiny strips parallel to each side in turn,
then the weight of each tiny strip acts at its midpoint. Therefore G, the
point at which the weight of the whole triangle acts, must lie on each
median. Therefore the medians must intersect.
If you are asked to show that the medians intersect at a point 2/3 of the
way down each of them then there is a short cut method you can use.
Let OG = g and OP = p.
g = p + 1/3 PA but p = b +
1/2BC = b + 1/2(-b + c) = 1/2(b +
Also PA = -p + a = a - 1/2(b +
So g = 1/2(b + c) + 1/3(a - 1/2(b
+ c)) = 1/3(a + b + c).
We see that g depends equally on a, b and c and
is therefore independent of the choice of median. Therefore it lies on
each median and gives their point of intersection.
return to the vectors homepage