The mid-points of the sides of a triangle are (1, 5, -1),(0,4,-2) and (2, 3, 4). Find its vertices.

Explanation

Let the vertices of `DeltaABC` are `A(x_(1),y_(1).z_(1)), B (x_(2).y_(2).z_(2)) and C (x_(3).y_(3),z_(3))`.
Since the mid - point of side BC is D (1,5-1)
then, `(x_(2)+x_(3))/2=1Rightarrowx_(2)+x_(3)=2`
`(y_(2)+y_(3))/2=5Rightarrowy_(2)+y_(3)=10`
`(z_(2)+z_(3))/2=-1Rightarrowz_(2)+z_(3)=-2`
Similarly, the mid-points of AB and AC are F ( 2,3,4) and E ( 0,4,-2)
`(x_(1)+x_(2))/2=2Rightarrowx_(1)+x_(2)=4`
`(y_(1)+y_(2))/2=3Rightarrowy_(1)+y_(2)=6`
` (z_(1)=z_(2))/2=4Rightarrowz_(1)+z_(2)=8`
`(x_(1)+x_(3))/2=0Rightarrowx_(1)+x_(3)=0`
`(y_(1)+y_(3))/2=4Rightarrowy_(1)+y_(3)=8`
`(z_(1)+z_(3))/2-2Rightarrowz_(1)+z_(3)=-4`
From Eqs. (i) and (iv)
`x_(1) + 2x_(2)+x_(3)=6`
from Eqs. (ii) and (v)
`y_(1)2y_(2)+y_(3)=16`
from Eqs. (iii) and (vi)
`z_(1) +2x_(2)+z_(3)=6`
from Eqs. (vii) and (x) ,
`2x_(2)=6Rightarrowx_(2)=3`
`x_(2)=3"then" x_(3)=-1`
`x_(3)=-1`
then `" " x_(1)=1Rightarrowx_(1)=1,x_(2)=3,x_(2)=-1`
from Eqs. (viii) and (xi)
`2y_(2)=8Rightarrow y_(2)=4`
`y_(2)=4`
`y_(1) =2`
`y_(1)=2`
`y_(3)=6`
`y_(1)=2,y_(2)=4,y_(3)=6`
form Eqs. (ix) and (xii)
`2z_(2)=10Rightarrowz_(2)=5`
`z_(2)=5`
`z_(1)=3`
`z_(1)=3`
`z_(3)=-7`
`z_(1)=3,z_(2)=5,z_(3)=-7`
So, the vertices of the trinagle A (1,2,3) B(3,4,5) and C (-1,6,-7)
Hence , centroid of the triangle `G((1+3-1)/3, (2+4+6)/3,(3+5-7)/3)` i.e. G ( 1,4,1//3)`