I would do this by comparing the gradient of z {(7, P), where P is the Bath Profit} with the tangent vectors to the constraints pointing at the optimum point: (1,0) from b≤10 and (-7,11) from 11a+7b≤100. So long as the gradient forms less than a right angle with these tangent vectors, it is pointing into the same corner as before:
(1,0) • (7,P) > 0 and (-7,11) • (7,P) > 0.
I'm too lazy to look up "slack variables" (the above in disguise?), but the above method works so long as as everything is linear. However, for more than 2 dimensions, those tangent vectors are the vectors along the 1-dimensional edges between vertices: intersections of several constraints.