Question

已知:直线方程为(2+m)x+(1-2m)y+4-3m=0,过定点M作直线L,使夹在两坐标之间的线段被点M平分,求直线L

Answer 1

(2+m)x+(1-2m)y+4-3m=0

(2x+y+4)+m(x-2y-3)=0

令2x+y+4=0

x-2y-3=0

联立解得:

x=-1,y=-2

所以:M(-1,-2)

所以:

直线L与X轴交于(-2,0),代入(2+m)x+(1-2m)y+4-3m=0,得:M=0

直线L的方程2x+y+4=0