Let O be the origin, A (1, 0) and B (0, 1) and P (x, y) are points such that xy > 0 and x+y<1, then

Since xy > 0, P either lies in the first quadrant or in the third quadrant. The inequality x + y < 1 represents all points below the line x + y = 1. So that xy>0 and x+y<1 imply that either P lies inside the triangle OAB or in the third quadrant.