The points \[A \left( x_1 , y_1 \right) , B\left( x_2 , y_2 \right) , C\left( x_3 , y_3 \right)\] are the vertices of ΔABC .
(i) The median from A meets BC at D . Find the coordinates of the point D.
(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.
(iii) Find the points of coordinates Q and R on medians BE and CF respectively such thatBQ : QE = 2 : 1 and CR : RF = 2 : 1.
(iv) What are the coordinates of the centropid of the triangle ABC ?
Solution
(i) Median AD of the triangle will divide the side BC in two equal parts.
Therefore, D is the midpoint of side BC.
Coordinates of D are \[\left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right)\]
(ii)THe point P divided the side AD in the ratio 2: 1.
Coordinates of P are \[\left( \frac{2 \times \left( \frac{x_2 + x_3}{2} \right) + 1 \times x_1}{2 + 1}, \frac{2 \times \left( \frac{y_2 + y_3}{2} \right) + 1 \times y_1}{2 + 1} \right) = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)\]
(iii)
Median BE of the triangle will divide the side AC in two equal parts.
Therefore, E is the midpoint of side AC.
Coordinates of E are \[\left( \frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2} \right)\] The point Q divided the side BE in the ratio 2: 1.
Coordinates of Q are \[\left( \frac{2 \times \left( \frac{x_1 + x_3}{2} \right) + 1 \times x_2}{2 + 1}, \frac{2 \times \left( \frac{y_1 + y_3}{2} \right) + 1 \times y_2}{2 + 1} \right) = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)\]
Similarly, Coordinates of Q are R are \[\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)\]
(iv)
The points P, Q and R coincides and is the centroid of the triangle ABC.
So, coordinates of the centroid is \[\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)\]
