Q.
The mid-points of the sides of a triangle are (5,7,11),(0,8,5) and (2,3,−1), then the vertices are
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Introduction to Three Dimensional Geometry
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Solution:
Let the vertices of a triangle are A(x1,y1,z1), B(x2,y2,z2) and C(x3,y3,z3)
Since, D,E and F are the mid-points of AC,BC and AB. ∴(2x1+x2,2y1+y2,2z1+z2)=(0,8,5) ⇒x1+x2=0,y1+y2=16,z1+z2=10....(i) ⇒(2x2+x3,2y2+y3,2z2+z3)=(2,3,−1) ⇒x2+x3=4,y2+y3=6,z2+z3=−2.....(ii(
and x2+x3=4,y2+y3=6,z2+z3=−2 ⇒(2x1+x3,2y1+y3,2z1+z3)=(5,7,11) ⇒x1+x3=10,y1+y3=14,z1+z3=22....(iii)
On adding Eqs. (i), (ii) and (iii), we get 2(x1+x2+x3)=14,2(y1+y2+y3)=36, 2(z1+z2+z3)=30 ⇒x1+x2+x3=7,y1+y2+y3=18 z1+z2+z3=15....(iv)
On solving Eqs. (i), (ii), (iii) and (iv), we get x3=7,x1=3,x2=−3 y3=2,y1=12,y2=4
and z3=5,z1=17,z2=−7
Hence, vertices of a triangle are (7,2,5),(3,12,17) and (−3,4,−7).
