Coordinate Geometry and Probability

[miteshsholay]

Points (5,0) , (0,0) and (0,10) form a triangle
Point (x,y) is a point within the triangle.
What is the probability that y<x?

OA [spoiler] 1/8 [/spoiler]

[Anurag@Gurome]

the point in the triangle with y < x will be the triangle formed with y = x with our triangle and which is shaded brown in the figure.
we need to get the point on the line formed by (0,10) and (5,0) where y = x meets.
the equation of the line formed by (0,10) and (5,0) => 2x + y = 10
the point of intersection 2x + y = 10 and x = y is (10/3, 10/3)
so the probability that y < x = area of the shaded region / area of the triangle
area of the shaded region = 1/2 * 5 * 10/3
area of the triangle = 1/2 * 5 * 10
the probability that y < x = 1/3

[GMATGuruNY]

Points (5,0) , (0,0) and (0,10) form a triangle
Point (x,y) is a point within the triangle.
What is the probability that y<x?

If (x,y) is within triangle BCO, then y<x.

Let CD = n.
Triangle BCD is similar to triangle ACO.
Thus, the legs in each triangle must be in the same ratio.
Since CO/OA = 5:10 = 1:2, CD:BD = n : 2n.
Thus, BD = 2n.

The line y=x forms a 45 degree angle with the x-axis.
Thus, in triangle BDO, BD = DO = 2n.
Thus, CO = n+2n = 3n.

Since the length of OA (10) is twice the length of CO (5), OA = 2(3n) = 6n.
Thus, BD/OA = 2n/6n = 1/3.

Since triangles BCO and ACO have the same base (CO), and the height of triangle BCO (BD=2n) is 1/3 the height of triangle ACO (OA=6n), the area of triangle BCO is 1/3 the area of triangle ACO.
Thus, P(y<x) = (area of BCO)/(area of ACO) = 1/3.

[miteshsholay]

thanks for the solutions. even i was getting the same ans, but got confused when saw a different official answer. may be its wrong.