2.S=9/2

3. Let a finite point P exist so that P(a, 0):

When P is a right-angled vertex, as shown in the figure: C is CF⊥x axis is F;

∫Rt△BOP∽Rt△PFC，

∴BO:PF=OP:CF，

That is 1/4-a=a/3,

A2-4a+3=0 and a= 1 or a = 3；;

The coordinate of point p is (1, 0) or (3,0).

To sum up, there are two P*** eligible.