If the coordinates of the mid-points of the sides of a triangle are (3, 4), (4, 6) and (5, 7), find its vertices.
Let A(x1 , y1), B(x2 , y2) and C(x3 , y3) be the vertices of triangle.
Let D(3, 4), E(4, 6) and F(5, 7) be the midpoints of sides BC, CA and AB respectively.
By midpoint formula.
x = , y =
For midpoint D(3, 4) of side BC,
3 = , 4 =
∴ = 6 and = 8 …(1)
For midpoint E(4, 6) of side CA,
4 = , 6 =
∴ = 8 and = 12 …(2)
For midpoint F(5, 7) of side AB,
5 = , 7 =
∴ = 10 and = 14 …(3)
Adding 1,2 and 3, we get,
=6 + 8 + 10
And = 8 + 12 + 14
∴ 2(= 24 and 2() = 34
∴ = 12 and = 17
+ 6 = 12 and + 8 = 17 …from 1
∴ = 6 and = 9
Substituting above values in 3,
6 + and 9 + = 14
∴ = 4 and = 5
Similarly for equation 2,
6 + = 8 and 9 + = 12
∴ = 2 and = 3
Hence the vertices of triangle are A(6 , 9), B(4 ,5) and C(2 ,3)