∵点D(3"," 3/2)在直线BC上,

　　∴3/a+3/2b=1,

　　∴6b+3a=2ab.0②

　　由①②,得2a2-21a+54=0,

　　即(2a-9)(a-6)=0,解得a=9/2或a=6.

　　因此,所求直线BC在两坐标轴上的截距为

　　{■(a=9/2@b=9/2)┤,或{■(a=6@b=3)┤.

　　∴直线BC的方程为2x/9+2y/9=1或x/6+y/3=1,

　　即2x+2y-9=0或x+2y-6=0.