Answer:

Let the coordinates of vertices A, B and C of the ∆ABC be (x1, y1), (x2, y2) and (x3, y3) respectively. Let P (3, 4), Q (4, 1) and R (2, 0) be the mid parts of sides BC, AC and AB respectively

Adding equations (1), (3) and (5)

2 (x1 + x2 + x3) = 18

⇒ x1 + x2 + x3 = 9 … (7)

Subtracting equation (1), (3) and (5) from equation (7), it is obtained

x1 = 9 – 6 = 3

x2 = 9 – 8 = 1

x3 = 9 – 4 = 5

Adding equations (2), (4) and (6)

2 (y1 + y2 + y3) = 10

⇒ y1 + y2 + y3 = 5 … (8)

Subtracting equation (2), (4) and (6) from equation (8), it is obtained

y1 = 5 – 8 = –3

y2 = 5 – 2 = 3

y3 = 5 – 0 = 5

Thus, the vertices of the triangles are (3, –3), (1, 3) and (5, 5).