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# 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

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