题目内容
如图,在平面直角坐标系中,Rt△OAB的直角边OA在x轴的正半轴上,点B在第象限,将△OAB绕点O按逆时针方向旋转至△OA′B′,使点B的对应点B′落在y轴的正半轴上,已知OB=2,
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(1)求点B和点A′的坐标;
(2)求经过点B和点B′的直线所对应的一次函数解析式,并判断点A是否在直线BB′上。
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(1)求点B和点A′的坐标;
(2)求经过点B和点B′的直线所对应的一次函数解析式,并判断点A是否在直线BB′上。
(1)B (
,1),A′(
,
);(2)在
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试题分析:(1)已知是直角三角形,并给出边和角,可先求得A,B点的坐标,进而根据旋转变换的特点,画图得出A′点的坐标;
(2)已知两点,根据待定系数法可以求出解析式,至于点A是否在直线上只需把点代入所求解析式,判断是否符合即可.
(1)在△OAB中,
∵
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∴AB=OB·
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OA= OB·
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∴点B的坐标为(
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过点A´作A´D垂直于y轴,垂足为D
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在Rt△OD A´中
DA´=OA´·
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OD=OA´·
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∴A´点的坐标为(
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(2)点B的坐标为(
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∴
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当
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∴A´(
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点评:本题主要错误在于一些学生在写点坐标时,纵坐标与横坐标调错,导致计算错误或在求一次函数的解析式时错误.
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