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Let A(x 1 , y 1 ), B(x 2 , y 2 ) and C(x 3 , y 3 ) be the vertices of ABC.
Let D(1, 2), E(0, -1) and F(2, -1) be the mid-point of sides BC, CA and AB respectively.
Since, D is the mid-point of BC.

Therefore, and
x 2 + x 3 = 2 and y 2 + y 3 = 4 (i)
Similarly, E and F are the mid-points of CA and AB respectively.

Therefore, = 0 and = -1
x 1 + x 3 = 0 and y 1 + y 3 = -2 (ii)

and, and
x 1 + x 2 = 4 and y 1 + y 2 = -2 (iii)
From equations (i), (ii) and (iii), we get,
(x 2 + x 3 ) + (x 1 + x 3 ) + (x 1 + x 2 ) = 2 + 0 + 4
and (y 2 + y 3 ) + (y 1 + y 3 ) + (y 1 + y 2 )
= 4 - 2 - 2
x 1 + x 2 + x 3 = 3
and y 1 + y 2 + y 3 = 0 (iv)
From equation (i) and (iv), we get,
x 1 + 2 = 3 and y 1 + 4 = 0
x 1 = 1 and y 1 = -4.
So, the co-ordinates of A are (1, -4).
From equations (ii) and (iv), we get,
x 2 = 3 and y 2 - 2 = 0
x 2 = 3 and y 2 = 2
So coordinates of B are (3, 2)
Using (iii) and (iv), we get,
4 + x 3 = 3, -2 + y 3 = 0
So, co-ordinates of C are (-1, 2)
Hence, the vertices of the triangle ABC are A(1, -4), B(3, 2) and C(-1, 2).

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