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Lets say that we have a 2 dimensional space and have a point $P(x, y)$. If I have an area, lets say the area that is inside these points

$P_1(-1,-1);P_2(-1,1);P_3(1,-1);P_4(1,1);$

and the point $P(\frac{1}{2},\sqrt{2})$,

how can I prove that the point $P$ is inside or outside the area?

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1 Answer 1

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If you mean by the inside the convex hull of those points, a point $P$ is in the inside of the polytope (convex hull of the points) $P_i$, $i=1,..n$, if and only if there are $\lambda_i$, $i=1,..,n$, $0 \leq \lambda_i$, $\sum_{i=1}^n \lambda_i=1$ and $$ P = \sum_{i=1}^n \lambda_i P_i $$ since the definition of the convex hull of those points is $$ \{ \sum_{i=1}^n \lambda_i P_i\ |\ \sum_{i=1}^n \lambda_i \}. $$ However, calculating it like this is hard. There is another way computing the so called supporting hyperplanes of the polytope and its inner normals. Have a look at Ewald: Combinatorial Convexity for more about this topic.

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