問題詳情:
已知直線l1:x+my+6=0,l2:(m-2)x+3y+2m=0,求m的值,使得:
(1)l1與l2相交; (2)l1⊥l2; (3)l1∥l2; (4)l1,l2重合.
【回答】
解 (1)由已知1×3≠m(m-2),即m2-2m-3≠0,
解得m≠-1且m≠3.
故當m≠-1且m≠3時,l1與l2相交.
(2)當1·(m-2)+m·3=0,即m=時,l1⊥l2.
(3)當1×3=m(m-2)且1×2m≠6×(m-2)或m×2m≠3×6,即m=-1時,l1∥l2.
(4)當1×3=m(m-2)且1×2m=6×(m-2),即m=3時,l1與l2重合.
知識點:直線與方程
題型:解答題