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).

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