If the coordinates of the midpoint …

Let A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) be the vertices of {tex}\triangle{/tex}ABC.

Let D (1, 2), E (0, -1), and F (2, -1) be the mid-points of sides BC, CA and AB respectively.
Since D is the mid-point of BC
{tex}\therefore \quad \frac { x _ { 2 } + x _ { 3 } } { 2 } = 1 \text { and } \frac { y _ { 2 } + y _ { 3 } } { 2 } = 2{/tex}
{tex}\Rightarrow{/tex}x₂ + x₃ = 2 and y₂ + y₃ = 4.............(i)
Similarly, E and F are the mid-points of CA and AB respectively.
{tex}\therefore \quad \frac { x _ { 1 } + x _ { 3 } } {2 } = 0 \text { and } \frac { y _ { 1 } + y _ { 3 } } { 2 } = - 1{/tex}
{tex}\Rightarrow{/tex}x₁ + x₃ = 0 and y₁ + y₃ = -2...........(ii)
and, {tex}\frac { x _ { 1 } + x _ { 2 } } { 2 } = 2 \text { and } \frac { y _ { 1 } + y _ { 2 } } { 2 } = - 1{/tex}
{tex}\Rightarrow{/tex}x₁ + x₂ = 4 and y₁ + y₂ = -2........(iii)
From (i), (ii) and (iii), we get
(x₂ + x₃) + (x₁ + x₃) + (x₁ + x₂)= 2 + 0 + 4:
and, (y₂ + y₃) + (y₁ + y₃) + (y₁ + y₂) =4 - 2 -2
{tex}\Rightarrow{/tex} 2(x₁ + x₂ + x₃) = 6 and 2 (y₁ + y₂ + y₃) = 0........(iv)
{tex}\Rightarrow{/tex}x₁ + x₂ + x₃ = 3 and y₁ + y₂ + y₃ = 0
From (i) and (iv), we get
x₁ +2 = 3 and y₁ + 4 = 0
{tex}\Rightarrow{/tex} x₁ = 1 and y₁ = -4
So, the coordinates of A are (1,-4)
From (ii) and (iv), we get
x₂ + 0 = 3 and y₂ - 2 = 0
{tex}\Rightarrow{/tex}x₂ = 3 and y₂ = 2
So, coordinates of B are (3, 2)
From (iii) and (iv), we get
x₃ + 4 = 3 and y₃ - 2 = 0
{tex}\Rightarrow{/tex}x₃ = -1 and y₃=2
So, coordinates of C are (-1, 2)
Hence, the vertices of the triangle ABC are A (1, -4), B (3, 2) and C (-1,2).