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

Explanation

Let `A(x_(1), y_(1),z_(1)), B(x_(2),y_(2),z_(2))and C(x_(3), y_(3), z_(3))` be the vertices of the given triangle, and let
D(1, 2, -3), E(3, 0, 1) and F(-1, 1, -4) be the mid-points of the sides AB, AC and BC respectively.
Now D is the mid-point of AB
`rArr(x_(1)+x_(2))/2=1, (y_(1)+y_(2))/2=2,(z_(1)+z_(2))/2=-3`
`rArr x_(1) +x_(2) = 2 …(i)`
`rArr y_(1) + y_(2) = 4 …(ii)`
and `z_(1) + z_(2) = -6 …(iii)`
E is the mid-point of AC
`rArr(x_(1)+x_(3))/2=3, (y_(1)+y_(3))/2=0,(z_(1)+z_(3))/2=1`
`rArr x_(1)+x_(3)=6 ...(iv)`
`y_(1)+y_(3)=0 …(v)`
and `z_(1) + z_(3)=2 ...(vi)`
F is the mid- point of BC
`rArr (x_(2)+x_(3))/2=-1, (y_(2)+y_(3))/2=1, (z_(2)+z_(3))/2=-4`
`rArr x_(2)+x_(3)=-2 …(vii)`
`y_(2)+y_(3)=2 …(viii)`
and `z_(2)+z_(3)=-8 …(ix)`
On adding (i), (iv), and (vii), we get ___
`2(x_(1)+x_(2)+x_(3))=6`
`rArr x_(1)+x_(2)+x_(3)=3 ...(x)`
On solving (i), (iv), and (vii) with (x), we get
`x_(1)=5, x_(2)=-3, x_(3)=1` On adding (ii), (v) and (viii), we get
`2(y_(1)+y_(2)+y_(3))=6`
`rArr y_(1) + y_(2) + y_(3) = 3 ` ...(xi)
On solving (ii), (v) and (viii) with (xi), we get
`y_(1)=1, y_(2)= 3, y_(3)=-1`
On adding (iii), (vi) and (ix), we get
`2(z_(1) + z_(2)+z_(3))=-12`
`rArr z_(1)+z_(2)+z_(3)=-6` … (xii)
On solving (iii), (vi) and (xi) with (xii), we get
`z_(1)=2, z_(2)=-8, z_(3)=0`
Hence, the vertices of the triangle are A(5, 1, 2), B(-3, 3, -8) and C(1, -1, 0).