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Weighted average of arbitrary k points from a line

  1. Oct 26, 2012 #1
    Suppose a set of k arbitrary points, x_i, 1<=i<=k, x_i from R^2 are selected from a line. How can it be shown that a weighted barycenter x_o=(o_i*x_i)/(o_1+o_2+...+o_k) also belongs to that line (assume o_i are arbitrary weights)? Does the choice of weights restrict the solutions (ie, a particular choice to satisfy that x_o is within the 'convex hull' of other points)?
     
  2. jcsd
  3. Oct 27, 2012 #2

    chiro

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    Hey onako.

    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).
     
  4. Oct 27, 2012 #3
    Let's simplify the question. Suppose that non-negative arbitrary weights o_i are associated with each x_i chosen from the straight line. Does x_o lie on that same line? The point is: points x_i are chosen as arbitrary points from the line, and are associated coefficients o_i which are non-negative.
     
  5. Oct 27, 2012 #4

    chiro

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    Are you looking at the triangle simplex in R^2?
     
  6. Oct 27, 2012 #5
    I'm concerned with a 2D case, with a straight line.
     
  7. Oct 27, 2012 #6

    chiro

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    If this line is in R^2 then they need a common gradient: try setting up a simple line based on y - y0 = m(x - x0) where m = (y1 - y0)/(x1 - x0) and showing that the gradient has a common form.

    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.
     
  8. Oct 29, 2012 #7
    A line in the plane satisfies a vector equation of the form
    [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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