beginner's complex!

Algebra Level pending

If l $z_{1}$ +m $z_{2}$ +n $z_{3}$ =0 and l+m+n=0 then find area of the triangle with $z_{1}$ , $z_{2}$ , $z_{3}$ as its vertices. (where l,m,n are real numbers)

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2 solutions

A Former Brilliant Member
Mar 17, 2015

$\displaystyle l=-(m+n) \\ \Rightarrow z_{1}= \dfrac{mz_{2}+nz_{3}}{m+n}$

Hence , we conclude that $z_{1}$ is a vector which divides the vectors $z_2 , z_3$ in the ratio m:n , and hence they all are collinear .

$\therefore$ We conclude that the triangle formed by the 3 vectors has area equal to 0 .

Yes that's the way I did it :) I don't think their is a need to edit it because that would make this simple problem much more simpler!

Prashant Kr - 6 years, 2 months ago

Raushan Singh
Mar 16, 2015

$z_{1}$ $z_{2}$ $z_{3}$ are collinear.

@Prashant Kr

Can you edit the question to specify what $z_{i} , i=1,2,3$ are ?

And yes , @Raushan Singh , what did you assume the z's to be, vectors or complex numbers ?

A Former Brilliant Member - 6 years, 2 months ago

I assumed it to be complex number. It won't make a difference if we assume it something else or would it? one can even use a,b,c instead of $z_{1}$ $z_{2}$ $z_{3}$

Raushan Singh - 6 years, 2 months ago

I agree with you . Btw I had taken them to be vectors (no big diff.) , if you want take a look at my solution . It's too simple actually .

A Former Brilliant Member - 6 years, 2 months ago

beginner's complex!

2 solutions

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