How to find corners / vertices of Polyhedron

Question

Polyhedron $\mathcal{P}\subseteq\mathbb{R}^2$, given by the inequalities:
$x_1+2x_2\geq 1,\qquad x_1\geq -1,\qquad x_1-x_2\geq -3,\qquad x_2\geq 1,\qquad -2x_1-x_2\leq 0 $
Giving matrix $A$ and vector $b$, such that $Ax\leq b$:
$$ A= \begin{pmatrix} -1 & -2\\ -1 & 0\\ -1 & 1\\ 0 & -1\\ -2&-1 \end{pmatrix}\qquad,\qquad b=\begin{pmatrix} -1\\ 1\\ 3\\ -1\\ 0\\ \end{pmatrix} $$

How do I proceed to find the corners/vertices of $\mathcal{P}$ ? Thanks for the help

Answer 1

The line $x_1+2x_2=1$ splits the $(x_1,x_2)$-plane into two halfplanes. In one of these we have $x_1+2x_2\geq1$. Draw the line on graphed paper, and mark the "good side" with an arrow $\nearrow$ standing orthogonally on the line, or similar. Do this for each of the five conditions. The set ${\cal P}$ consists of all the points that are on the "good side" of every drawn line. You will have to compute some intersection points.