(2)解方程组{■(x+y+2=0"," @2x+2y+3=0"," )┤0├ ■("①" @"②" )┤

①×2-②得1=0,矛盾,方程组无解.

所以直线l1与l2无公共点,即l1∥l2.

(3)解方程组{■(x"-" y+1=0"," @2x"-" 2y+2=0"," )┤0├ ■("①" @"②" )┤

①×2得2x-2y+2=0.

因此①和②可以化为同一个方程,即①和②表示同一条直线,所以直线l1与l2重合.

9.求经过两条直线2x-3y-3=0和x+y+2=0的交点,且与直线3x+y-1=0平行的直线l的方程.

解:由方程组{■(2x"-" 3y"-" 3=0"," @x+y+2=0)┤解得{■(x="-" 3/5 "," @y="-" 7/5 "." )┤

因为所求直线l和直线3x+y-1=0平行,

所以直线l的斜率为-3.根据点斜式可得

y-("-" 7/5)=-3[x"-" ("-" 3/5)],

即所求直线方程为15x+5y+16=0.

二、能力提升

1.若三条直线2x+3y+8=0,x-y-1=0和x+by=0相交于一点,则b等于(　　)

A.-2 B.-1/2 C.2 D.1/2

解析:由{■(2x+3y+8=0"," @x"-" y"-" 1=0"," )┤解得{■(x="-" 1"," @y="-" 2"," )┤代入x+by=0,得-1-2b=0,b=-1/2.

答案:B

2.已知直线kx-y+1=3k,则当k变化时,所有直线都通过定点(　　)

A.(0,0) B.(0,1) C.(3,1) D.(2,1)

解析:当k=0时,直线方程为-y+1=0.当k=1时,直线方程为x-y+1=3.解方程组{■(x"-" y+1=3"," @"-" y+1=0"," )┤得{■(x=3"," @y=1"," )┤即直线都通过定点(3,1).