Question

已知直线L1:kx-y-2k=0,L2:(2k-1)x-2ky-2=0 (1)证明直线L1过定点 (2)若L1垂直于L2

Answer 1

1)L1化为：y=k(x-2)

当x=2时,无论k为何值,都有y=0

所以过定点(2,0)

2)L1的斜率=k

L2的斜率=(2k-1)/(2k)

两者垂直,则斜率的积为-1

即k(2k-1)/(2k)=-1

2k-1=-2

k=-1/2

得L2: -2x+y-2=0