(2) Let E and F be the midpoint of AB and CD respectively, and point P be on AD, connecting PE and PF, and extending the X axis to M and N, then the area of triangle △PMN is equal to the area of rectangular ABCD.

As shown in figure P(5/2kloc-0/), when the midpoint p (5/2,8) of AD is taken, there is an isosceles triangle △PMN, and PM=PN=√89.

Because m (-5/2,0), the slope of the straight line PM is K=(0-8)/(-5/2-5/2)=8/5.

As shown in Figure (2), when point p (3,8) is taken, there is an isosceles triangle △PMN, MP=MN= 10.

Because m (-3,0), the slope k of the straight line PM = (0-8)/(-3-3) = 4/3.

As shown in Figure (3), when point p (2,8) is taken, there is an isosceles triangle △PMN, NP=NM= 10.

Because m (-2,0), the slope k of the straight line PM = (0-8)/(-2-2) = 2.