The points \(A(x_1,y_1),\ B(x_2,y_2)\) and \(C(x_3,y_3)\) are the vertices of \(\triangle ABC\).
(i) The median from \(A\) meets \(BC\) at \(D\). Find \(D\).
(ii) Find the point \(P\) on \(AD\) such that \(AP:PD=2:1\).
(iii) Find points \(Q\) and \(R\) on medians \(BE\) and \(CF\) respectively such that \(BQ:QE=2:1\) and \(CR:RF=2:1\).
(iv) Hence, write the coordinates of the centroid of \(\triangle ABC\).
(i) \(D\bigl(\dfrac{x_2+x_3}{2},\;\dfrac{y_2+y_3}{2}\bigr)\).
(ii) \(P\bigl(\dfrac{x_1+x_2+x_3}{3},\;\dfrac{y_1+y_2+y_3}{3}\bigr)\).
(iii) \(Q\) and \(R\) are also \(\bigl(\dfrac{x_1+x_2+x_3}{3},\;\dfrac{y_1+y_2+y_3}{3}\bigr)\).
(iv) Centroid \(G\bigl(\dfrac{x_1+x_2+x_3}{3},\;\dfrac{y_1+y_2+y_3}{3}\bigr)\).
The median from \(A\) meets side \(BC\) at its midpoint \(D\).
So \(D\) is the midpoint of \(B(x_2,y_2)\) and \(C(x_3,y_3)\):
\[ D\left(\dfrac{x_2+x_3}{2},\;\dfrac{y_2+y_3}{2}\right) \]
Point \(P\) lies on \(AD\) such that \(AP:PD = 2:1\). So \(P\) divides the segment joining \(A(x_1,y_1)\) and \(D\left(\dfrac{x_2+x_3}{2},\dfrac{y_2+y_3}{2}\right)\) internally in the ratio \(2:1\).
Using the section formula:
\[ P = \left(\dfrac{2x_D + 1x_A}{3},\;\dfrac{2y_D + 1y_A}{3}\right) \]
Substitute \(x_D=\dfrac{x_2+x_3}{2},\; y_D=\dfrac{y_2+y_3}{2}\):
\[ P = \left(\dfrac{2\cdot\dfrac{x_2+x_3}{2} + x_1}{3},\;\dfrac{2\cdot\dfrac{y_2+y_3}{2} + y_1}{3}\right) \]
Simplifying:
\[ P\left(\dfrac{x_1+x_2+x_3}{3},\;\dfrac{y_1+y_2+y_3}{3}\right) \]
First find the midpoints of the other two sides.
Midpoint of \(AC\): Let this be \(E\).
\[ E\left(\dfrac{x_1+x_3}{2},\;\dfrac{y_1+y_3}{2}\right) \]
Midpoint of \(AB\): Let this be \(F\).
\[ F\left(\dfrac{x_1+x_2}{2},\;\dfrac{y_1+y_2}{2}\right) \]
Point Q on median BE: \(Q\) divides \(BE\) internally in the ratio \(BQ:QE = 2:1\).
Using section formula between \(B(x_2,y_2)\) and \(E\left(\dfrac{x_1+x_3}{2},\dfrac{y_1+y_3}{2}\right)\):
\[ Q = \left(\dfrac{2x_E + 1x_B}{3},\;\dfrac{2y_E + 1y_B}{3}\right) \]
Substitute \(x_E=\dfrac{x_1+x_3}{2},\; y_E=\dfrac{y_1+y_3}{2}\):
\[ Q = \left(\dfrac{2\cdot\dfrac{x_1+x_3}{2} + x_2}{3},\;\dfrac{2\cdot\dfrac{y_1+y_3}{2} + y_2}{3}\right) = \left(\dfrac{x_1+x_2+x_3}{3},\;\dfrac{y_1+y_2+y_3}{3}\right) \]
Point R on median CF: \(R\) divides \(CF\) internally in the ratio \(CR:RF = 2:1\).
Using section formula between \(C(x_3,y_3)\) and \(F\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\):
\[ R = \left(\dfrac{2x_F + 1x_C}{3},\;\dfrac{2y_F + 1y_C}{3}\right) \]
Substitute \(x_F=\dfrac{x_1+x_2}{2},\; y_F=\dfrac{y_1+y_2}{2}\):
\[ R = \left(\dfrac{2\cdot\dfrac{x_1+x_2}{2} + x_3}{3},\;\dfrac{2\cdot\dfrac{y_1+y_2}{2} + y_3}{3}\right) = \left(\dfrac{x_1+x_2+x_3}{3},\;\dfrac{y_1+y_2+y_3}{3}\right) \]
The centroid is the common point of intersection of all three medians.
Since \(P\), \(Q\), and \(R\) all have the same coordinates, this common point is the centroid \(G\):
\[ G\left(\dfrac{x_1+x_2+x_3}{3},\;\dfrac{y_1+y_2+y_3}{3}\right) \]
Final Note: The centroid coordinates are simply the average of the x-coordinates and the average of the y-coordinates of the three vertices.