已知两直线:L1:(3+M)X+4Y=5-3M L2:2X+(5+M)Y=8 当M为何值时L1与L2平行垂直相交

题目

已知两直线:L1:(3+M)X+4Y=5-3M L2:2X+(5+M)Y=8 当M为何值时L1与L2平行垂直相交
请用高一必修2 直线与方程的有关知识解答.

答案

L1:(3+M)X+4Y=5-3M L2:2X+(5+M)Y=8 L1:(3+M)X+4Y=5-3M ==>y=〔5-3M -(3+M)X〕/4 L2:2X+(5+M)Y=8 ==>y=(8-2x)/(5+m) 1.相交 〔5-3M -(3+M)X〕/4=(8-2x)/(5+m) m≠-1,-7 2.平行 k1=k2,-(3+M)/4=-2/(5+m) m=-1,-7 3.垂...