Checking if a point is an extreme point of the convex hull of a set of 4 points

Question

Given the set $D = \text{conv}(\{ (1, 2, 2), (-1, 2, 3), (15, -2, 0), (\frac{15}{2}, 0, \frac{5}{4})\})$. How can I check if $x = (\frac{15}{2}, 0, \frac{5}{4})$ is an extreme point of $D$?

I have trouble visualizing how the set looks like. So far I have only checked that $x$ cannot be represented by a convex combination of $(1, 2, 2)$ and $(-1,2,3)$, $(1, 2, 2)$ and $(15,-2,0)$, or $(-1,2,3)$ and $(15,-2,0)$. However, I'm not sure if it is enough the conclude that $x$ is not a extreme point of $D$.

Answer 1

You can check if your 4 points are affinely independent, which is equivalent to checking the (linear) independence of the differences $b-a$, $c-a$, $d-a$ for some fixed $a$. If that is the case, then the representation of your simplex as a convex hull is unique, and all the given points are its vertices, hence extreme points.