Question

A (4, 2), В (6, 5) and C (1, 4) are the vertices of ΔABC.

1. The median from A meets BC in D. Find the coordinates of the point D. 
2. Find the coordinates of point P on AD such that AP : PD = 2 : 1. 
3. Find the coordinates of the points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1. 
4. What do you observe?

Answer 1

Given: A(4, 2), B(6, 5), C(1, 4).

Step-wise calculation:

1. (i) D is midpoint of BC:

`x_D = (6 + 1)/2 = 7/2`

`y_D = (5 + 4)/2 = 9/2`

So `D = (7/2, 9/2)`.

2. (ii) P on AD with AP : PD = 2 : 1 → `t = 2/(2+1) = 2/3` from A toward D. 

`D - A = (7/2 - 4, 9/2 - 2)`

= `(-1/2, 5/2)`

`P = A + (2/3)(D - A)`

= `(4 - 1/3, 2 + 5/3)`

= `(11/3, 11/3)`

3. (iii) Medians: E = midpoint of AC, F = midpoint of AB.

`E = ((4 + 1)/2, (2 + 4)/2)`

= `(5/2, 3)` 

`F = ((4 + 6)/2, (2 + 5)/2)`

= `(5, 7/2)`

Q on BE with BQ : QE = 2 : 1 → `t = 2/3` from B: 

`E - B = (5/2 - 6, 3 - 5)`

= `(-7/2, -2)` 

`Q = B + (2/3)(E - B)`

= `(6 - 7/3, 5 - 4/3)`

= `(11/3, 11/3)`

R on CF with CR : RF = 2 : 1 → `t = 2/3` from C: 

`F - C = (5 - 1, 7/2 - 4)`

= `(4, -1/2)` 

`R = C + (2/3)(F - C)`

= `(1 + 8/3, 4 - 1/3)`

= `(11/3, 11/3)`

P = Q = R = `(11/3, 11/3)`. These coincide at the centroid of triangle ABC also equal to the average of the vertices: `((4 + 6 + 1)/3, (2 + 5 + 4)/3) = (11/3, 11/3)`.