1
设D点(x,y)
则：AD=(x,y)-(1,2)=(x-1,y-2)
BC=(-1,5)-(0,3)=(-1,2)
AD⊥BC,故：AD·BC=(x-1,y-2)·(-1,2)
=1-x+2y-4=0
即：x-2y=-3-----------------(1)
BD=(x,y)-(0,3)=(x,y-3)
DC=(-1,5)-(x,y)=(-1-x,5-y)
BD、DC共线,即：
x/(x+1)=(y-3)/(y-5)
即：2x+y=3-----------------(2)
由(1)、(2)得：x=3/5,y=9/5
即：AD=(3/5-1,9/5-2)=(-2/5,-1/5)
2
BD=(x,y-3)=(3/5,-6/5)=3(1,-2)/5
DC=(-1-x,5-y)=(-8/5,16/5)=-8(1,-2)/5
即：BD=-3DC/8
即：BC=DC-DB=DC+BD
=-5BD/3
即：BD=-3BC/5
故：AD=BD-BA=-BA-3BC/5