题目内容
在平面直角坐标系中,点O为坐标原点,已知点A(0,1)、点B(
,0),则∠OAB = ,点G为△ABO重心,则点G的坐标是 .
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516535311.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408230125165511645.png)
先根据题意画出图形,直接根据锐角三角函数的定义即可求出∠OAB的度数;再根据直角三角形的性质求出AB的长,进而得出AB的中线OD的长,由三角形重心的性质得出OG的长,再根据AB两点的坐标求出其中点D的坐标,利用待定系数法求出OD的解析式,设出G点坐标,利用两点间的距离公式即可得出G点坐标.
解答:解:如图所示:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408230125165673487.png)
∵A(0,1),B(
,0),
∴OA=1,OB=
,
∴tan∠OAB=
=
,
∴∠OAB=60°;
∴∠ABO=30°,
∴AB=2OA=2,
∵点D为AB的中点,G为重心,
∴OD=
AB=
×2=1,OG=
OD=
×1=
,D(
,
),
设过O、D两点的直线解析式为y=kx,则
=
k,解得k=
,
∴过O、D两点的直线解析式为y=
x,
∴设G(x,
x),则OG=
,解得x=
或x=-
(舍去),
∴G(
,
).
故答案为:60°;(
,
).
解答:解:如图所示:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408230125165673487.png)
∵A(0,1),B(
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516582344.png)
∴OA=1,OB=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516582344.png)
∴tan∠OAB=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516613696.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516582344.png)
∴∠OAB=60°;
∴∠ABO=30°,
∴AB=2OA=2,
∵点D为AB的中点,G为重心,
∴OD=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516676338.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516676338.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516707382.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516707382.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516707382.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516738453.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516676338.png)
设过O、D两点的直线解析式为y=kx,则
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516676338.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516738453.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516801419.png)
∴过O、D两点的直线解析式为y=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516801419.png)
∴设G(x,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516801419.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516847939.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516801419.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516801419.png)
∴G(
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516801419.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516894327.png)
故答案为:60°;(
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516801419.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012516894327.png)
![](http://thumb.zyjl.cn/images/loading.gif)
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