Answer :

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)


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