Choose your points, go up from the y intercept to a chosen x point. Then go over the the x point. Up and over is the formula I use.
Answer:
Option 4 : 
Step-by-step explanation:
<u>See the attached figure:</u>
To find the vertices of the feasible region, we need to graph the constraints, then find the area included by them, then calculate the vertices which is the intersection between each two of them.
As shown, the shaded area represents the solution of the constraints
So, the vertices of the feasible region are:

Ugh, more algebra pretending to be geometry. Why?
This is a pretty awful instance of the genre, with no units on the constant 16, assumed to be degrees.
The total arc measure in a circle is 360 degrees. Geometry part over.
3m + 16 + 2m + 3m = 360
8m = 344
m = 344/8 = 43
Arc measures:
Arc 3m+16 = 3(43)+16 = 145°. The angle subtended by that arc is half that, 145/2=72.5°
Arc 2m=2(43)=86°. The angle subtended is half that, 43°
Arc 3m=3(43)=129°. The angle subtended is half that, 64.5°
212= 2 X 2 X 53
212= 4X 53