Degeneracy Condition

I understood that when plotting the feasible area there had to be an intersection with more than two lines.

In the case of:

$$\text{Max } z=2x_1+x_2$$ S.T $$\begin{cases} 4x_1+3x_2\leq 12\\ 4x_1+x_2 \leq 8\\ 4x_1-x_2 \leq 8\\ x_1,x_2\geq 0 \end{cases}$$

The plot is which mean that there is almost intersection of the there functions, is it still degeneracy?

Edit: yes this LP admits degenerate solutions, e.g., $$(x_1, x_2)=(2,0)$$, since the constraints $$4x_1+x_2 \leq 8, 4 x_1-x_2 \leq 8, x_2 \geq 0$$ are all active at this point, and $$3$$ active constraints in a 2d space implies degeneracy.
• Ah right, I missed the non-negativity constraints. Yes, $4x_1+x_2 \leq 8, 4 x_1-x_2 \leq 8, x_2 \geq 0$ are all active at $(x_1, x_2)=(2,0)$, so yes that point is degenerate. – Ryan Cory-Wright Apr 17 '19 at 20:44
• In your plot, don't $x=0, 4x+y=8, 4x-y=8$ intersect at the same point? – Ryan Cory-Wright Apr 18 '19 at 14:12