Make shapes with inequality

Question

The following set : $$\{(x,y) \in \mathbb{R}^2 \mid x+y \leq 1, x\geq 0, y \geq 0 \}$$ is a triangle. One way to see it is simply that we draw all points under the line of equation $y = 1-x$ with positive coordinates.

My question is :

• Is it possible with inequalities (just as the one that describe a triangle) to draw some other nice shapes like parallelogram or more generally regular polygons ?

• Moreover I suspect that there is some linear algebra behind these inequalities. So maybe for example linear algebra can help proving that the above inequality makes a triangle.

Thank you !

Answer 1

Yes. In fact, every convex polygon can be described as

$$\{(x,y)| a_1x + b_1 y\leq c_1, a_2x+b_2x\leq c_2,\dots, a_nx + b_nx \leq c_n\}$$

or, in linear algebra terms:

For every convex polygon $P$, there exists some matrix $A\in\mathbb R^{n\times 2}$ and a vector $c\in\mathbb R^n$ such that $$P=\left\{(x,y)| A\begin{bmatrix}x\\y\end{bmatrix}\leq c\right\}$$