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Suppose a line is given in 2D, and a set of $k$ arbitrary points $x_1, x_2, \dots, x_k$ along that line. For some non-zero weights $w_i$ associated with each $x_1$, it can be shown that a weighted barycenter $y=\frac{w_ix_i}{\sum_i w_i}$ also lies on that line. What are the constraints on such weights for, eg, the point $y$ to lie 'outside the convex hull' of points $x_i$? I'm interested in general constraints on such weights with respect to point $y$ position relative to $x_i$.

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I think question is not completely clear to me so excuse me if I misunderstood. Since points lie on line, any well-defined linear combination would also lie on same line. Hence as long as y is defined, it will be on that line. Also, convex-hull of all x_{i} is simply the line segment you may call it 1-simplex and to lie inside it the only constraint is min-max of x_{i}, like you would have done to check if C lied between A and B, given A,B,C were collinear.

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Ok. I'm interested in the constraints on weights to make the linear combination lie within the convex hull. What about all weights being positive, and sum to 1? –  user506901 Oct 27 '12 at 13:20
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I think there are no simple constraints on $w_i$ to judge if a point is inside the convex hull of $n$ points $p_1,\ldots,p_n$, except for simple ($n\leq3$) cases.

If you have $n>3$ points, and given a specific point $p$, the solution of weights $w_i$ are not unique. So I think in general there are no simple condition.

However, if you have triangle $p_1,p_2,p_3$, then assume the weights sum to 1, then you can solve $w_1,w_2,w_3$ uniquely. Then if a point $p$ lies inside the triangle, then all the weights must be non-negative.

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