BD=BC+CD =(a+b)+(3a-5b) =4a-4bA、B、D三点共线设可设AB=tBD2a+kb=t(4a-4b)(2-4t)a+(k+4)b=02-4t=0 ;k+4=0解得k=-4 ,t=1/2
AB=2a+kbBD=BC+CD=4a-4b若A、B、D三点共线，则AB||BD则AB/BD=常数只有k=-2
