Question

两条直线y=kx+2k+1和x+2y-4=0的交点在第四象限，求k的取值范围

Answer 1

y = kx + 2k + 1
x + 2y - 4 = 0
解此方程组得：
x = (2 - 4k)/(2k + 1)
y = (6k + 1)/(2k + 1)
因为交点在第四象限
所以 x > 0 , y < 0
所以 (2 - 4k)/(2k + 1) > 0 , (6k + 1)/(2k + 1) < 0
所以 -1/2 < k < -1/6

Answer 2

求交点坐标：
∵y=kx+2k+1和x+2y-4=0
∴x+2*(kx+2k+1)-4=0
(1+2k)x+4x-2=0
x=(2-4k)/(1+2k)
y=k*(2-4k)/(1+2k)+2k+1
∵交点在第四象限
∴x>0，y<0
(2-4k)/(1+2k)>0
[k*(2-4k)/(1+2k)]+2k+1<0

(4k-2)/(1+2k)<0
-1/2<k<1/2

[k*(2-4k)/(1+2k)]+2k+1<0
(6k+1)/(2k+1)<0
-1/2<k<-1/6

综合：-1/2<k<-1/6

Answer 3

k<-1/2 <-1/6