Question

Prove that the coordinates of the centroid of the triangle whose vertices are $(x_1,y_1),(x_2,y_2)and(x_3,y_3)$ are $(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$ and also, deduce that the medians of a triangles are concurrent.

Answer 1

Let $A(x_1,y_1),B(x_2,y_2)andC(x_3,y_3)$ be the vertices of ∆ABC whose medians are AD, BE and CF respectively so, D, E and F are respectively the mid- points of BC ,CA and AB.

Coordinates of D are $\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}$

Coordinates of a point dividing AD in the ratio 2:1 are

$(1.x_1+2\frac{\frac{x_2+x_3}{2}}{1+2},1.y_1+2\frac{\frac{y_2+y_3}{2}}{1+2})$

$=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

The coordinates of E are $(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2})$

The coordinates of a point dividing BE in the ratio 2:1 are

$(\frac{\frac{1.x_2+2(x_1+x_3)}{2}}{1+2},\frac{\frac{1.y_2+2(y_1+y_3)}{2}}{1+2})$

$=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

Similarly the coordinates of appoint dividing CF in the ratio 2:1 are

$(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

Thus , BE and CF divides them in the ratio 1:2 .