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Question

The mid -points of the sides of a triangle are(1,51),(0,4,2)and(2,3,4). Find its vertices. 
 


  1. A(1,2,3),B(3,4,5)andC(1,6,7).

  2. A(1,2,3),B(3,4,5)andC(2,6,7).

  3. A(1,2,3),B(5,4,5)andC(1,6,7).

  4. 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=1x2+x3=2,y2+y3=10,z2+z3=2
E is the mid point of CA
x1+x32=0,y1+y32=4,z1+z32=2x1+x3=0,y1+y3=8,z1+z3=4
F is the mid point of AB
x1+x22=2,y1+y22=3,z1+z22=4x1+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+4x1+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)=25+8z1+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).
 

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