故直线(a-3)x+2ay+6=0恒过定点(2,-1).

又点(2,-1)在第四象限,所以该直线恒过第四象限.

答案:D

5.若直线x-y-1=0和x-ky=0的交点在第三象限,则k的取值范围是(　　)

A.0

C.k>1 D.k<0

解析:由{■(x"-" y"-" 1=0"," @x"-" ky=0"," )┤解得{■(x=k/(k"-" 1) "," @y=1/(k"-" 1) "." )┤由交点在第三象限,得 k/(k"-" 1)<0,k-1<0,解得0

答案:B

6.已知两条直线l1:ax+3y-3=0,l2:4x+6y-1=0.若l1与l2相交,则实数a满足的条件是　　　　　.

解析:由题意得6a-12≠0,即a≠2.

答案:a≠2

7.经过原点和直线l1:x-3y+4=0与l2:2x+y+5=0的交点的直线方程为　　　　　.

解析:由{■(x"-" 3y+4=0"," @2x+y+5=0"," )┤得交点坐标为("-" 19/7 "," 3/7),

所以所求方程为y=-3/19 x,即3x+19y=0.

答案:3x+19y=0

8.判断下列各对直线的位置关系,若相交,求出交点坐标:

(1)l1:2x+y+3=0,l2:x-2y-1=0;

(2)l1:x+y+2=0,l2:2x+2y+3=0;

(3)l1:x-y+1=0,l2:2x-2y+2=0.

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

所以直线l1与l2相交,交点坐标为(-1,-1).