已知直线L1:(a+1)x+y-(a+1)/(a*a+1)=0,L2:x-y-(a*a-3)/(a*a+1)=0
问题描述:
已知直线L1:(a+1)x+y-(a+1)/(a*a+1)=0,L2:x-y-(a*a-3)/(a*a+1)=0
(1)当a为何值时,l1//l2?当a为何值时,l1⊥l2?
(2)若L1与l2相交,且交点在第一象限,求a的取值范围.
答
1、(a+1)/1=1/-1
a=-2
2、(a+1,1)·(1,-1)=0
a+1-1=0
a=0
3、交点:
x0=(a-1)(a+2)/(a+2)(a^2+1)
∵相交,a≠-2
∴x0=(a-1)/(a^2+1)
y0=-(a-2)(a+1)/(a^2+1)
x0>0
得a>1
y0>0
得a>2或a1