The mid-pointsof the sides of the …

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a given triangle as shown in figure.

Now, (3,4) is the mid-point of AB, therefore,
{tex}3=\frac{x_1+x_2}{2}\ {/tex}and {tex}4=\frac{y_1+y_2}{2}{/tex}
x₁ + x₂ = 6 and y₁ + y₂ = 8 ..... (i)
(2,0) is the mid-point of BC, then,
{tex}2=\frac{x_2+x_3}{2}{/tex} and {tex} 0=\frac{y_2+y_3}{2}{/tex}
x₂ + x₃ = 4 and y₂ + y₃ = 0 ..........(ii)
(4,1) is the mid-point of AC, then,
{tex}4=\frac{x_1+x_3}{2}{/tex} and {tex}1=\frac{y_1+y_3}{2}{/tex}
x₁ + x₃ = 8 and y₁ + y₃ = 2 .........(iii)
Subtracting (ii) from (iii), we get,
x₁ - x₂ = 4 and y₁ - y₂ = 2 ........ (iv)
Adding (i) and (iv), we get,
2x₁ = 10 and 2y₁ = 10
x₁ = 5 and y₁ = 5
From (i), we have,
x₂ = 6 - 5 = 1 and y₂ = 8 - 5 = 3
From (ii), we have,
x₃ = 4 - 1 = 3 and y₃ = 0 - y₂ = 0 - 3 = -3
Thus (5, 5), (1, 3) and (3, -3) are the vertices of triangle.