A.-12 B.-10 C.10 D.12

解析由l2:x+y-4=0,l3:2x-y+1=0,可得交点坐标为(1,3),代入直线l1:ax+2y+6=0,可得a+6+6=0,∴a=-12.故选A.

答案A

6.已知直线ax+by-2=0,且3a-4b=1,则该直线必过定点　　　　　.

解析由3a-4b=1,得b=(3a"-" 1)/4,代入ax+by-2=0,得a(4x+3y)=y+8.令{■(4x+3y=0"," @y+8=0"," )┤解得{■(x=6"," @y="-" 8"." )┤

答案(6,-8)

7.导学号57084072已知直线l被两直线l1:4x+y+6=0,l2:3x-5y-6=0截得的线段的中点是原点O,则直线l的方程为　　　　　.

解析由已知可设直线l的方程为y=kx,

　　由{■(y=kx"," @4x+y+6=0"," )┤得x=-6/(k+4).

　　由{■(y=kx"," @3x"-" 5y"-" 6=0"," )┤得x=6/(3"-" 5k).

　　由已知可得-6/(k+4)+6/(3"-" 5k)=0,

　　解得k=-1/6,故所求直线l的方程为y=-1/6x,

　　即x+6y=0.当斜率不存在时,不合题意.

答案x+6y=0

8.已知直线x+y-3m=0和2x-y+2m-1=0的交点M在第四象限,求实数m的取值范围.

解由{■(x+y"-" 3m=0"," @2x"-" y+2m"-" 1=0"," )┤得{■(x=(m+1)/3 "," @y=(8m"-" 1)/3 "." )┤

　　故交点M的坐标为(m+1)/3, (8m"-" 1)/3.

　　交点M在第四象限,

　　得{■((m+1)/3>0"," @(8m"-" 1)/3<0"," )┤解得-1

　　故m的取值范围是-1,1/8.

9.导学号57084073过点M(0,1)作直线,使它被两直线l1:x-3y+10=0,l2:2x+y-8=0所截得的线段恰好被M所平分,求此直线方程.

解法一过点M且与x轴垂直的直线显然不合题意,故可设所求直线方程为y=kx+1.

　　设所求直线与已知直线l1,l2分别交于A,B两点.

由{■(y=kx+1"," @x"-" 3y+10=0"," )┤得A的横坐标xA=7/(3k"-" 1).