If the coordinates of the mid-points of the sides of a triangle are (1, 1), (2, -3) and (3, 4), find the vertices of the triangle.
Let A(x1 , y1), B(x2 , y2) and C(x3 , y3) be the vertices of triangle.
Let D(1, 1), E(2, -3) and F(3, 4) be the midpoints of sides BC, CA and AB respectively.
By midpoint formula.
x = 
, y = 
For midpoint D(1, 1) of side BC,
1 = 
, 1 = 
∴ 
= 2 and 
= 2 …(1)
For midpoint E(2, -3) of side CA,
2 = 
, -3 = 
∴ 
= 4 and 
= -6 …(2)
For midpoint F(3, 4) of side AB,
3 = 
, 4 = 
∴ 
= 6 and 
= 8 …(3)
Adding 1,2 and 3, we get,
= 2 + 4 + 6
And 
= 2 – 6 +8
∴ 2(
= 12 and 2(
) = 2
∴ 
= 6 and 
= 1
+ 2 = 6 and 
+ 2 = 2 …from 1
∴ 
= 4 and 
= 0
Substituting above values in 3,
4 + 
and 0 + 
= 8
∴ 
= 2 and 
= 89
Similarly for equation 2,
4 + 
= 6 and 0 + 
= -6
∴ 
= 2 and 
= -6
Hence the vertices of triangle are A(4 , 0), B(2 ,8) and C(0 ,-6)