设m∈R,求两条直线L1:x+my+6=0与L2:(m-2)x+3y+2m=0 交点P的轨迹方程.
问题描述:
设m∈R,求两条直线L1:x+my+6=0与L2:(m-2)x+3y+2m=0 交点P的轨迹方程.
答
两条直线L1:x+my+6=0与L2:(m-2)x+3y+2m=0相交
则x+my+6=(m-2)x+3y+2m
(m-3)x+(3-m)y+2m-6=0
(m-3)(x-y+2)=0
当m=3时,两直线重合,交点P方程即为x+3y+6=0
当m≠3时,交点P方程为x-y+2=0