# Medians of a Triangle are Concurrent

Proof the medians of a triangle are concurrent using co-ordinate geometry.

To proof this theorem we need to use the formula of co-ordinates of the point dividing the line segment joining two given points in a given ratio and the mid-point formula.

Let (x₁, y₁), (x₂, y₂) and (x₃, y₃) be the rectangular cartesian co-ordinates of the vertices M, N and O respectively of the triangle MNO. If P, Q and R be the mid-points of the sides NO, OM and MN respectively, then the co-ordinates of P, Q and R are ((x₂ + x₃)/2, (y₂ + y₃)/2)) , ((x₃ + x₁)/2 , (y₁ + y₂)/2)) respectively.

Now, we take a point G₁ on the median MP such that MG₁, G₁P = 2 : 1. Then the co-ordinates of G₁ are

= ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3)



Again, we take a point G₂ on the median NQ such that NG₂ : G₂Q = 2 : 1. Then the co-ordinates of G₂ are

= ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3)

Finally, we take a point G₃ on the median OR such that OG₃ : G₃R = 2 : 1. Then the co-ordinates of G₃ are

= {(x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3}

Thus we see that G₁, G₂ and G₃ are the same point. Hence, the medians of the triangle are concurrent and at the point of concurrence the medians are divided in the ratio 2 : 1.

Note:

The point of concurrence of the medians of the triangle MNO is called its centroid and the co-ordinates of the centroid are {(x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3}

Worked-out examples on medians of a triangle are concurrent:

1. If the Co-ordinates of the three verticals of a triangle are (-2, 5), (-4, -3) and (6, -2), find the Co-ordinates of the centroid of the triangle.

Solution:

The Co-ordinates of the centroid of the triangle formed by the joining the given points are {(- 2 - 4 + 6)/3}, (5 - 3 - 2)/3)}

[Using the formula {(x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3}]

= (0, 0).

2. The co-ordinates of the vertices A, B, C of the triangle ABC are (7, -3), (x, 8) and (4, y) respectively; if the co-ordinates of the centroid of the triangle be (2, -1), find x and y.

Solution:

Clearly, the co-ordinates of the centroid of the triangle ABC are

{(7 + x + 4)/3, (- 3 + 8 + y)/3)} = {(11 + x)/3, (5 + y)/3}.

By problem, (11 + x)/3 = 2

or, 11 + x = 6

or x = -5

And (5 + y)/3 = -1

or, (5 + y) = -3

or, y = -8.

Therefore, x = -5 and y = -8

3. The co-ordinates of the vertex A of the triangle ABC are (7, -4). If the co-ordinates of the centroid of the triangle be (1, 2), find the co-ordinates of the mid-point of the side BC.

Solution:

Let G (1, 2) be the centroid of the triangle ABC and D (h, k) be the mid-point of the side BC.

Since G (1, 2) divides the median AD internally in the ratio 2 : 1, hence we must have,

(2 ∙ h + 1 ∙ 7)/(2 + 1) = 1

or, 2h + 7 = 3

or, 2h = -4

or, h = -2

And {2 ∙ k + 1 ∙ (-4)}/(2 + 1) = 2

or, 2k - 4 = 6

or, 2k = 10

or, k = 5.

Therefore, the co-ordinates of the mid-point of the side BC are (-2, 5).

Co-ordinate Geometry

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