Finding the centroid of a triangle using complex numbers

Baartzy89
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Hi all,

I'm preparing for a deferred exam this semester after falling ill last year. Just looking over my course notes and have a question. I understand how this works in the big picture scheme. What I don't understand however is how my instructor simplified the original equation.

1. Homework Statement

Step 1) (1-t)z1 + t*(z2+z3/2) = (1-s)z1 + s(z2+z3/2)

Step 2) Simplifies to;
(2-s)z1+(t-2+2s)z2+(t-s)z3 = 0

Since z1, z2 and z3 aren't collinear, their coefficients in this equation must be zero. Therefore we have;
a) 2-s-2t = 0
b) t-2+2s = 0
c) t-s = 0

Then we readily find t = s = 2/3

Which is then substituted into the original equation for medians to find that it equals (z1+z2+z3)/3

Homework Equations


I understand how this works in the big picture scheme. What I don't understand however is how my instructor simplified the original equation step 1 to step 2.[/B]

The Attempt at a Solution


My attempt from equating the equations;
(1-t)z1 + t*(z2+z3/2) = (1-s)z1 + s(z2+z3/2)

0 = (2-s)z1+(t-2+2s)z2+(t-s)z3
= (1-s)z1 + s(z1+z3/2) + s(z2+z3/2) - (1-t)z1 - s(z2+z3/2)
= z1 - s*z1 - z1 + t*z1 - s*(z2/2) - s*(z3/2) + s*(z2/2) + s*(z3/2)
= t*z1 - s*z1

Therefore t*z1 = s*z1 and divide both sides by z1 t = s

I feel this comes out slightly like my instructors, but its faulty somewhere...
 
Physics news on Phys.org
Why not expand the brackets in the first equation, multiply through by 2, then group like terms?
 

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