(1) AB=(4,1)、BC=(x,y)、CD=(-2,-2)
AC=AB+BC
=(x+4,y+1)
AD=AC+CD
=(x+4-2,y+1-2)
=(x+2,y-1)
BC∥AD
x(y-1)-y(x+2)=0
xy-x-xy-2y=0
y=-x/2
(2) BD=AD-AB
=(x+2-4,y-1-1)
=(x-2,y-2)
AC⊥BD
(x+4)(x-2)+(y+1)(y-2)=0
x^2+2x-8+y^2-y-2=0
x^2+2x+(-x/2)^2-(-x/2)-10=0
5/4x^2+5/2x-10=0
5x^2+10x-40=0
5(x^2+2x-8)=0
5(x+4)(x-2)=0
x1=-4
x2=2
y1=-(-4)/2=2
y2=-2/2=-1
x=-4时，y=2
x=2时，y=-1
