Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of point D.
(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.
(iii) Find the coordinates of point Q and R on medians BE and CF respectively such that BQ : QE = 2:1 and CR : RF = 2 : 1.
(iv) What do you observe?
[Note: The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2 : 1.]
(v) If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of triangle ABC, find the coordinates of the centroid of the triangle.
(i) Coordinates of D = ( (6+1)/2, (5+4)/2 ) = (7/2, 9/2)
So, D is (7/2, 9/2)
(ii) Coordinates of P can be calculated as follows:
Coordinates of P = ( [2(7/2) + 1(4)]/(2 + 1), [2(9/2) + 1(2)]/(2 + 1) ) = (11/3, 11/3)
So, P is (11/3, 11/3)
(iii) Coordinates of E can be calculated as follows:
Coordinates of E = ( (4+1)/2, (2+4)/2 ) = (5/2, 6/2) = (5/2 , 3)
So, E is (5/2 , 3)
Point Q and P would be coincident because medians of a triangle intersect each other at a common point called centroid.
Coordinates of Q =( [2(5/2) + 1(6)]/(2 + 1), [2(3) + 1(5)]/(2 + 1) ) = (11/3, 11/3)
F is the mid- point of the side AB
Coordinates of F = ( (4+6)/2, (2+5)/2 ) = (5, 7/2)
Point R divides the side CF in ratio 2:1
Coordinates of R = ( [2(5) + 1(1)]/(2 + 1), [2(7/2) + 1(4)]/(2 + 1) ) = (11/3, 11/3)
(iv) Coordinates of P, Q and R are same which shows that medians intersect each other at a common point, i.e. centroid of the triangle.
(v) If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of triangle ABC, the coordinates of centroid can be given as follows:
x = (x1 + x2 + x3)/3 and y = (y1 + y2 + y3)/3