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- Thread starter onako
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- #2

chiro

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If this is a straight line then from this you have n-independent coeffecients (since the o_i's are independent) and since you can specify those then you can choose the coeffecients such that these match the ones of the original line.

If all the x_i's line on a straight-line, then provided that you can pick the o_i's, then you can solve for the particular values of o_i to provide equal coeffecients.

The only thing though about this is that you are dividing by the sum of the all the o_i's which means you lose a degree of freedom and the only way you can get a solution is if you have an extra constraint on the o_i's and this constraint is exactly what is imposed.

If your line is a unique line then the solution in terms of your o_i's should be unique as well since you have a constraint on the sum of the weights which means the scalar multiples should disappear as well (I'm consider scalar multiples of a linear equation).

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- #4

chiro

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Are you looking at the triangle simplex in R^2?

- #5

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I'm concerned with a 2D case, with a straight line.

- #6

chiro

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For the bary-centric case the sum of all the weights should be 1 so you can cancel that out, and see if calculating this line gives a common form in terms of the coeffecients.

- #7

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[itex] Nx=a [/itex]

So if each xi satisfies that equation, you just have to show that a weighted average of them also satisfies that equation. As long as the weights add up to 1 it should work.

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