题目内容
直线
与反比例函数
(x>0)的图像交于点A,与坐标轴分别交于M、N两点,当AM=MN时,求k的值.
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过点A作AB⊥x轴, 垂足为B,对于直线y=kx+
当x="0" 时.
即OM=
∵AM=MN
∴AN=2MN
∵Rt△MON ∽Rt△ABN
∴
∴
将
代入
中得 x=1
∴A(1,
)
∵点A在直线y=kx+
上
∴
= k+
∴k =
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当x="0" 时.
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即OM=

∵AM=MN
∴AN=2MN
∵Rt△MON ∽Rt△ABN
∴
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∴
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将
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∴A(1,
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∵点A在直线y=kx+
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∴
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∴k =
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过点A作AB⊥x轴,垂足为B,先求出M点的坐标得到OM=
;由AM=MN,易得OM为△ABN的中位线,根据中位线的性质得到AB=2MO=2
,得到A点的纵坐标为2
,然后将y=2
代入y=
中得x=1,则A点坐标为(1,2
),然后把A(1,2
)代入y=kx+
得到关于k的方程,再解方程即可.
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