In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a

In the xy- plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a point (x,y) is selected at random from the triangular region, What is the probability that x-y>0 ?

A. 1/5
B. 1/3
c. 1/2
D. 2/3
E. 4/5



We have right triangle with the area = 4*5/2 = 10. Consider the line y<x. All the points which satisfy this equation (are below the line y = x) and lie in the triangular region obviously will have x more than y, which is exactly what we want (as x > y --> x - y > 0).

The probability that the point will be from this region is: (Area of this region)/(Area of the triangle).

Favorable region is also right triangle with vertexes at (0,0) (0,4) and (4,4). As y = x intersects the side of our original triangle at the point (4,4). You''ll see it easily if you draw it. So favorable area = 4*4/2 = 8.

\(P=\frac{8}{10}=\frac{4}{5}\).

Answer: E.