(1)
证源手茄明:
(2+m)x+(1-2m)y+4-3m=0得
(2+m)*(x+1)+(1-2m)*(y+2)=0
所以直线恒过M(-1,-2)
(2)直线方程化为斜雹察截式
x/((3m-4)/(2+m))+y/((3m-4)/(1-2m))=1
(3m-4)/(2+m)<0
(3m-4)/(1-2m)<0
得S=1/2*(3m-4)/(2+m)*(3m-4)/(1-2m),(-2
得S≤0,薯辩或S≥4
显然S≥4
当S=4时,m=0
所以ΔAOB面积的最小值4
此时直线方程:y=-2x-4