Find the coordinates of the centroid of the triangles if points `D(-7,6),E(8,5)` and `F(2,-2)` are the midpoints of the sides of that triangle.

Explanation

Let `A(x_(1),y_(1)),B(x_(2),y_(2))` and `C(x_(3),y_(3))` be the vertices of `DeltaABC`. Let `D(-7,6),E(8,5)` and `F(2,-2)` be the midpoints of sides AB,BC and AC respectively.
Let `G(x,y)` be the centroid.
Now D is the midpoint of AB. ltBrgt `-7=(x_(1)+x_(2))/2`
`:.x_(1)=x_(2)=-7xx2`
`:.x_(1)+x_(2)=-14`.............1
Similarly
`x_(2)+x_(3)=16`......2
and `x_(1)+x_(3)=4`............3
adding 1,2 and 3 we get
`x_(1)+x_(2)+x_(2)+x_(3)+x_(1)+x_(3)=-14+16+4`
`:.2x_(1)+2x_(2)+2x_(3)=6`
`:.2(x_(1)+x_(2)+x_(3))=6`
`:.x_(1)+x_(2)+x_(3)=6/2`
`:.x_(1)+x_(2)+x_(3)=3`.............4
D is the midpoint of AB.
`:.` by midpoint formula,
`6=(y_(1)+y_(2))/2`
`:.y_(1)_y_(2)+12`......5
Similarly `y_(2)+y_(3)=10` .......6
`y_(1)+y_(3)=-4`......7
Adding 5, 6 and 7 we get
`y_(1)+y_(2)+y_(2)+y_(3)+y_(1)+y_(3)=12+10-4`
`:.2y_(1)+2y_(2)+2y_(3)=18`
`:.2(y_(1)+y_(2)+y_(3))=18`
`:.y_(1)+y_(2)+y_(3)=18/2`
`:.y_(1)+y_(2)+y_(3)=9`..............8
`G(x,y)` is the centroid of `DeltaABC`.
`:.` by centroid formula,
`x=(x_(1)+x_(2)+x_(3))/3,y=(y_(1)+y_(2)+y_(3))/3`
`:.x=3/3`....[From 4], `:.y=9/3` ..[From 8]
`:.x=1`, :.y=3`
The coordinates of point G are (1,3).