# Coordinates of Centroid

We shall show that the medians of a triangle are concurrent and shall find the coordinates of the point of concurrence (called the centroid) in terms of the coordinates of the vertices

Let the vertices of the triangle be A(x_{1},y_{1}), B(x_{2},y_{2}) and C(x_{3},y_{3}).

Then D, the mid-point of BC is .

Let us find the point on AD which divides it in the ratio 2 : 1. Its coordinates are

that is Â Â Â Â Â Â Â Â Â --------- (1)

Again, the coordinates of the point on BE, which divides it in the ratio 2:1 are the same as (1).

Â

And the coordinates of the point on CF, which divides it in the ratio 2:1 are also the same as (1).

Hence point (1) is common to the three medians, which shows that the medians are concurrent.

The coordinates of the centroid, generally denoted by G, are given by .

Â

(i) We know that the coordinates of the centroid of triangle with vertices A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) is

Thus, centroid of triangle ABC is

Â

Two vertices of a triangle are (âˆ’1, 4) and (5, 2). If the centroid is (0, âˆ’3), find the third vertex.

Let the third vertex of the triangle be (x, y). Then centroid of triangle with vertices (âˆ’1, 4), (5, 2) and (x, y) is

But the centroid of triangle is given as (0, âˆ’3). Therefore

â‡’ 4 + x = 0 and 6 + y = âˆ’ 9 â‡’ x = âˆ’ 4 and y = âˆ’15.

Thus, the coordinates of the third vertex are (âˆ’4, âˆ’15).

The vertices of a triangle are (1, 2) (h, âˆ’3) (âˆ’4, k). Find the values of h and k if the centroid of the triangle be at the point (5, âˆ’1).

Â The centroid of the triangle with vertices (1, 2), (h, âˆ’3) and (âˆ’4, k) is

we are given centroid of the triangle is (5, âˆ’1).

Thus

If (âˆ’2, 3), (4, âˆ’3) and (4, 5) are the mid-points of the sides of a triangle, find the coordinates of the centroid of the triangle.

Let the coordinates of the vertices of the triangle be A(x_{1}, y_{1}), B(x_{2, }y_{2}) and C(x_{3}, y_{3}). Let D(âˆ’2, 3), E(4, âˆ’3), and F(4, 5) be the mid-points of BC, CA and AB, respectively.

Since D is mid-point of BC, coordinates D are

Thus __Â Â Â Â Â Â Â __(1)

Similarly, __Â Â Â Â Â Â Â Â Â __(2)

and __Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â __(3)

Adding the equations in (1), (2), and (3) we get

2(x_{1} + x_{2} + x_{3}) = 12, 2(y_{1} + y_{2} + y_{3}) = 10

â‡’ x_{1} + x_{2} + x_{3 }= 6, y_{1} + y_{2} + y_{3} = 5

As x_{2} + x_{3} = âˆ’4, y_{2} + y_{3} = 6, we get x_{1} = 10, y_{1} = âˆ’1

Similarly, x_{2} = âˆ’2, y_{2} = 11 and x_{3} = âˆ’2, y_{3} = âˆ’5.

Thus, the vertices of the triangle are (10, âˆ’1), (âˆ’2, 11) and (âˆ’2, âˆ’5).Â

The centroid of triangle ABC is