Sivaram has answered your title problem: you simply plot the points $(0,0)$, $(b,-a)$ and $(-b,a)$ or any two of them. Then you draw the straight line through these points, extending it if necessary.
Let's extend it in two ways to your optimisation problem. First, how to draw $ax+by=c$ where $c$ is not $0$. Plot the points $(\frac{c}{a},0)$ and $(0,\frac{c}{b})$; this also works if one of $a$ or $b$ is $0$. Then you draw the straight line through these points, extending it if necessary. That, together with Sivaram's advice, is enough to draw your five constraints. This will give you a feasible region which is a simple convex polygon, in this case with five vertices.
The second extension is to draw a line parallel to $ax+by=0$ which passes through the point $(d,e)$. The answer is to plot the points $(d,e)$, $(d+b,e-a)$ and $(d-b,e+a)$ or any two of them. Then you draw the straight line through these points, extending it if necessary. You can do this with your function to be maximised and each of the five points. One or more of the five resulting lines will be further to the top-right than the others. The point(s) in your feasible region this passes though will give where $z$ is maximised.
If after all this you do not find that the maximum possible value of $z$ is $62$ then either you or I have made a mistake.