Proof the medians of a triangle are concurrent using coordinate geometry.
To proof this theorem we need to use the formula of coordinates of the point dividing the line segment joining two given points in a given ratio and the midpoint formula.
Let (x₁, y₁), (x₂, y₂) and (x₃, y₃) be the rectangular cartesian coordinates of the vertices M, N and O respectively of the triangle MNO. If P, Q and R be the midpoints of the sides NO, OM and MN respectively, then the coordinates 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 coordinates 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 coordinates 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 coordinates 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 coordinates of the centroid are {(x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3}
Workedout examples on medians of a triangle are concurrent:
1. If the Coordinates of the three verticals of a triangle are (2, 5), (4, 3) and (6, 2), find the Coordinates of the centroid of the triangle.
Solution:
The Coordinates 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 coordinates of the vertices A, B, C of the triangle ABC are (7, 3), (x, 8) and (4, y) respectively; if the coordinates of the centroid of the triangle be (2, 1), find x and y.
Solution:
Clearly, the coordinates 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 coordinates of the vertex A of the triangle ABC are (7, 4). If the coordinates of the centroid of the triangle be (1, 2), find the coordinates of the midpoint of the side BC.
Solution:
Let G (1, 2) be the centroid of the triangle ABC and D (h, k) be the midpoint 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 coordinates of the midpoint of the side BC are (2, 5).
● Coordinate Geometry
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