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(2014•甘肃二模)已知点P在直线x+2y-1=0上,点Q在直线x+2y+3=0上,PQ的中点为M(x0,y0),且y0>x0+2,则y0x0的取值范围是()A.(−12,−15)B.(−12,−15]C.[−12,−15]D.[−12,−15)
题目详情
(2014•甘肃二模)已知点P在直线x+2y-1=0上,点Q在直线x+2y+3=0上,PQ的中点为M(x0,y0),且y0>x0+2,则
的取值范围是( )
A.(−
,−
)
B.(−
,−
]
C.[−
,−
]
D.[−
,−
)
| y0 |
| x0 |
A.(−
| 1 |
| 2 |
| 1 |
| 5 |
B.(−
| 1 |
| 2 |
| 1 |
| 5 |
C.[−
| 1 |
| 2 |
| 1 |
| 5 |
D.[−
| 1 |
| 2 |
| 1 |
| 5 |
▼优质解答
答案和解析
设P(x1,y1),
=k,则y0=kx0,∵PQ中点为M(x0,y0),∴Q(2x0-x1,2y0-y1)
∵P,Q分别在直线x+2y-1=0和x+2y+3=0上,
∴x1+2y1-1=0,2x0-x1+2(2y0-y1)+3=0,
∴2x0+4y0+2=0即x0+2y0+1=0,
∵y0=kx0,
∴x0+2kx0+1=0即x0=-
,
又∵y0>x0+2,代入得kx0>x0+2即(k-1)x0>2即(k-1)(-
)>2即
<0
∴-
<k<-
故选A
| y0 |
| x0 |
∵P,Q分别在直线x+2y-1=0和x+2y+3=0上,
∴x1+2y1-1=0,2x0-x1+2(2y0-y1)+3=0,
∴2x0+4y0+2=0即x0+2y0+1=0,
∵y0=kx0,
∴x0+2kx0+1=0即x0=-
| 1 |
| 1+2k |
又∵y0>x0+2,代入得kx0>x0+2即(k-1)x0>2即(k-1)(-
| 1 |
| 1+2k |
| 5k+1 |
| 2k+1 |
∴-
| 1 |
| 2 |
| 1 |
| 5 |
故选A
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