Answer :

Let A(x_{1} , y_{1}), B(x_{2} , y_{2}) and C(x_{3} , y_{3}) 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)

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