  Question

# The mid -points of the sides of a triangle are(1,5−1),(0,4,−2)and(2,3,4). Find its vertices.    A(1,2,3),B(3,4,5)andC(−1,6,−7). A(1,2,3),B(3,4,5)andC(−2,6,−7). A(1,2,3),B(−5,4,5)andC(−1,6,−7). A(1,2,3),B(3,4,5)andC(−1,−6,−7).

Solution

## The correct option is A A(1,2,3),B(3,4,5)andC(−1,6,−7). Let A(x1,y1,z1),B(x2,y2,z2)andC(x3,y3,z3) be the vertices of the given triangle and let D(1,5,−1),E(0,4,−2)andF(2,3,4) be the mid points of the sides BC, CA and AB respectively. Now, D is mid-point of BC ∴x2+x22=1,y2+y32=5,z2+z32=−1⇒x2+x3=2,y2+y3=10,z2+z3=−2 E is the mid point of CA ∴x1+x32=0,y1+y32=4,z1+z32=−2⇒x1+x3=0,y1+y3=8,z1+z3=−4 F is the mid point of AB ∴x1+x22=2,y1+y22=3,z1+z22=4⇒x1+x2=4,y1+y2=6,z+1+z2=8 Adding first three equations in (i) ,(ii) and (iii) , we obtain  2(x1+x2+x3)=2+0+4⇒x1+x2+x3=3 Solving first three equations in (i) , (ii) and (iii) with  x1+x2+x3=3, we obtain x1=1,x2=3,x3=−1. Adding next three equations in (i) (ii) and (iii), we obtain  2(y1+y2+y3)=10+8+6 ⇒y1+y2+y3=12 Solving next three equations in (i) , (ii) and (iii) with y1+y2+y3=12, we obtain y1=2,y2=4,y3=6. Adding last three equations in (i) ,(ii) and (iii) we obtain 2(z1+z2+z3)=−2−5+8⇒z1+z2+z3=1. Solving last three equation in (i) , (ii) and (iii) with z1+z2+z3=1, we obtain z1=3,z2=5,z3=−7. Thus the vertices of the triangle are A(1,2,3),B(3,4,5)andC(−1,6,−7).  Suggest corrections   