Finding Coordinates of the Centroid from Coordinates of the midpoints of the sides of a triangle

The coordinates of the midpoints of the sides of a triangle are (1, 2) (0, 1) and (2, 1), find its centroid.

In any triangle, medians are the three segments connecting the vertices of the triangle to the midpoints of the sides opposite to the vertices. It turns out the three medians in any triangle always intersect at a single point. That point is called the centroid of the triangle. 

The question is, can we find the coordinates of the centroid of a triangle from the given coordinates of the midpoints of the sides? 

Yes, we can. The definition of the centroid doesn’t really give us any clue in that direction. But there are some properties of the centroid that can help us out here.  

I don’t know of any formula or rule which directly relates the coordinates of the midpoints of the sides with the coordinates of the centroid. But I do know of a relation relating the coordinates of the centroid with the coordinates of the vertices of a triangle. The x and y coordinates of a centroid are simply the average of x and y coordinates of the three vertices. 

So if (x1, y1), (x2, y2) and (x3, y3) are the coordinates of the vertices of a triangle, the x and y coordinates of the centroid will be

(\(\frac{x_1 + x_2 + x_3}{3}\), \(\frac{y_1 + y_2 + y_3}{3}\)) 

But how do we find the coordinates of the vertices in our example? 

Well, the coordinates of the midpoints of the sides can help us with that. It’s a fairly simple but also a little tedious process. 

While the coordinates of the centroid are the average of the coordinates of the three vertices, x and y coordinates of the midpoint of a line segment are the average of the x and y coordinates of the two end points of the segment. 

In the above figure, 1 is the x coordinate of the midpoint of AB, and x coordinates of A and B are x1 and x2. So the average(or mean) of x1 and x2 is 1. 

\(\frac{x_1 + x_2}{2} = 1\)

 \(x_1 + x_2 = 2\) ..(1)

2 is the y coordinate of the midpoint of AB, and y1 & y2 are y coordinates of A and B. 

So, \(\frac{y_1 + y_2}{2} = 2\)

       \(y_1 + y_2 = 4\) ..(2)

Similarly, (2, 1) is the average of (x2, y2) & (x3, y3). 

        \(\frac{x_2 + x_3}{2} = 2\)

        \(x_2 + x_3 = 4\)..(3)

and,  \(\frac{y_2 + y_3}{2} = 1\)

        \(y_2 + y_3 = 2\)..(4)

And, (0, 1) is the average (x1, y1) & (x3, y3). 

        \(\frac{x_1 + x_3}{2} =0\)

        \(x_1 + x_3 = 0\)..(5)

and,  \(\frac{y_1 + y_3}{2} = 1\)

        \(y_1 + y_3 = 2\)..(6)

Six equations and six unknowns is what we end up with. We can solve a system of two equations with the same two unknowns, but in our case, it looks like no two equations have the same two unknowns. 

From Equation (5), 

x3 = - x1

Substituting for x3 in equation (3), we get

\(x_2 + (- x_1) = 4\)..(3)

\(x_2 - x_1 = 4\)..(3)

which now has the same two unknowns as equation (1).

\(x_2 + x_1 = 2\)..(1)

If we add the re-formed equation (3) to equation (1), this will get rid of the x1 unknown so we can easily solve for x2.

Adding (1) and (3),

\(x_2 + x_1 + x_2 - x_1 = 2 + 4\)

\(2 x_2 = 6\)

\(x_2 = 3\)

Substituting the value of x2 back in (1), we get

\(x_2 + x_1 = 2\)..(1)

\(3 + x_1 = 2\)

\(x_1 = 2 - 3 = -1 \)

Substituting -1 for x1 in (5),

 x3 = - x1 = - (-1) = 1

So we have found the x coordinates of all three vertices of the triangle. We can determine the y coordinates in the same way. 

From equation (6), 

\(y_1 + y_3 = 2\)..(6)

\(y_3 = 2 - y_1\)

Substituting for y3 in equation (4),

 \(y_2 + y_3 = 2\)..(4)

 \(y_2 + 2 - y_1 = 2\)

  \(y_2 = y_1\)

We can straightaway swap y2 for y1  in equation (2). 

\(y_1 + y_2 = 4\) ..(2)

\(y_1 + y_1 = 4\)

\(2 y_1 = 4\)

\(y_1 = \frac{4}{2} = 2\)

Since, \(y_2 = y_1\)

\(y_2 = 2\)

Substituting 2 for y2 back in (4)

\(y_2 + y_3 = 2\)..(4)

\(2 + y_3 = 2\)

\(y_3 = 0\)

Therefore, the coordinates of the vertices are 

        (x1, y1) = (-1, 2), 

        (x2, y2) = (3 , 2), 

and  (x3, y3) = (1 , 0)

The coordinates of the centroid will then be

(p, q) = (\(\frac{x_1 + x_2 + x_3}{3}\), \(\frac{y_1 + y_2 + y_3}{3}\))

   = (\(\frac{-1 + 3 + 1}{3}\), \(\frac{2 + 2 + 0}{3}\))

        = (\(\frac{3}{3}\), \(\frac{4}{3}\)) 

        = (1, \(\frac{4}{3}\))