已知:直線y=kx(k≠0)經(jīng)過點(3,-4).
(1)求k的值;
(2)將該直線向上平移m(m>0)個單位,若平移后得到的直線與半徑為6的⊙O相離(點O為坐標原點),試求m的取值范圍.
【答案】
分析:(1)中,因為直線y=kx(k≠0)經(jīng)過點(3,-4),所以把點的坐標直接代入即可求出k=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103194018743471838/SYS201311031940187434718016_DA/0.png)
.
(2)中,可設平移后的直線為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103194018743471838/SYS201311031940187434718016_DA/1.png)
x+m(m>0),則該直線與x軸、y軸的交點分別是A(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103194018743471838/SYS201311031940187434718016_DA/2.png)
m,0),B(0,m),即OA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103194018743471838/SYS201311031940187434718016_DA/3.png)
m,OB=m,利用勾股定理可求出AB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103194018743471838/SYS201311031940187434718016_DA/4.png)
m,過點O作OD⊥AB于D,運用△AOB的面積可求出AB上的高OD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103194018743471838/SYS201311031940187434718016_DA/5.png)
m,又因該直線與半徑為6的⊙O相離(點O為坐標原點),所以OD>6.從而可求出m>10.
解答:解:
(1)依題意得:-4=3k,
∴k=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103194018743471838/SYS201311031940187434718016_DA/6.png)
.(3分)
(2)由(1)及題意知,設平移后得到的直線l所對應的函數(shù)關系式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103194018743471838/SYS201311031940187434718016_DA/7.png)
x+m(m>0).(4分)
設直線l與x軸、y軸分別交于點A、B,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103194018743471838/SYS201311031940187434718016_DA/images8.png)
如右圖所示
當x=0時,y=m;當y=0時,x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103194018743471838/SYS201311031940187434718016_DA/8.png)
m.
∴A(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103194018743471838/SYS201311031940187434718016_DA/9.png)
m,0),B(0,m),即OA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103194018743471838/SYS201311031940187434718016_DA/10.png)
m,OB=m.
在Rt△OAB中,AB=
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103194018743471838/SYS201311031940187434718016_DA/12.png)
.(5分)
過點O作OD⊥AB于D,
∵S
△ABO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103194018743471838/SYS201311031940187434718016_DA/13.png)
OD•AB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103194018743471838/SYS201311031940187434718016_DA/14.png)
OA•OB,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103194018743471838/SYS201311031940187434718016_DA/15.png)
ODו
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103194018743471838/SYS201311031940187434718016_DA/16.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103194018743471838/SYS201311031940187434718016_DA/17.png)
ו
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103194018743471838/SYS201311031940187434718016_DA/18.png)
m•m,
∵m>0,解得OD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103194018743471838/SYS201311031940187434718016_DA/19.png)
m(6分)
∵直線與半徑為6的⊙O相離,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103194018743471838/SYS201311031940187434718016_DA/20.png)
m>6,解得m>10.
即m的取值范圍為m>10.(8分)
點評:此類題目是函數(shù)與圓的知識的綜合運用,難點在第(2)題,解決的根據(jù)是直線和圓相離?圓心到直線的距離大于圓的半徑.