If the coordinates of the mid-points of the sides of a triangle be (3, -2), (-3, 1) and (4, -3), then find the coordinates of its vertices.
Let A(x1 , y1), B(x2 , y2) and C(x3 , y3) be the vertices of triangle.
Let D(3, -2), E(-3, 1) and F(4, -3) be the midpoints of sides BC, CA and AB respectively.
By midpoint formula.
x = 
, y = 
For midpoint D(3, -2) of side BC,
3 = 
, -2 = 
∴ 
= 6 and 
= -4 …(1)
For midpoint E(-3, 1) of side CA,
-3 = 
, 1 = 
∴ 
= -6 and 
= 2 …(2)
For midpoint F(4, -3) of side AB,
4 = 
, -3 = 
∴ 
= 8 and 
=-6 …(3)
Adding 1,2 and 3, we get,
=6 -6 + 8
And 
= -4 + 2 -6
∴ 2(
= 8 and 2(
) = -8
∴ 
= 4 and 
= -4
+ 6 = 4 and 
– 4 = -4 …from 1
∴ 
= -2 and 
= 0
Substituting above values in 3,
-2 + 
and 0 + 
= -6
∴ 
= 10 and 
= -6
Similarly for equation 2,
-2 + 
= -6 and 0 + 
= 2
∴ 
= -4 and 
= 2
Hence the vertices of triangle are A(-2 , 0), B(10 ,-6) and C(-4 ,2)