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) Given, D is the median to BC. So, coordinates of D using distance formula are

\( x \ = \ \frac{6+1}{2} \ = \ \frac{7}{2} \) and \( y \ = \ \frac{5+4}{2} \ = \ \frac{9}{2} \)

(ii) Coordinates of P can be calculated using section formula

\( x \ = \ \frac{2( \frac{7}{2}) + 1(4)}{3} \ = \ \frac{11}{3} \) and \( y \ = \ \frac{2( \frac{9}{2}) + 1(2)}{3} \ = \ \frac{11}{3} \)

(iii) Coordinates of E
= ( \( \frac{4+1}{2} \ , \ \frac{2+4}{2} \) )
= ( \( \frac{5}{2} , 3 \) )

So, coordinates of Q are,
( \( \frac{2( \frac{5}{2}) + 1(6)}{3} \ , \ \frac{2(3) + 1(5)}{3} \) )
= ( \( \frac{11}{3} , \frac{11}{3} \) )

Coordinates of F
= ( \( \frac{4+6}{2} , \frac{2+5}{2} \) )
= ( \( 5, \frac{7}{2} \) )

So, coordinates of R are,
( \( \frac{2(5) + 1(1)}{3} , \frac{2 ( \frac{7}{2}) + 1(4)}{3} \) )
= ( \( \frac{11}{3} , \frac{11}{3} \) )

(iv) We observed that coordinates of P , Q, R are same i.e., ( \( \frac{11}{3} , \frac{11}{3} \) ) which shows that medians intersect each other at a common point which is called centroid of the triangle.

(v) If A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) are the vertices of triangle ABC, then the coordinates of the centroid of the triangle is

\( x \ = \ \frac{x_1 + x_2 + x_3}{3} \ and \ \frac{y_1 + y_2 + y_3}{3} \)