線段AB,其中點A(1,-4),點B(5,-4),將線段AB繞中點C逆時針旋轉30°后,得到新的線段A′B′,則線段A′B′的解析式為________.
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分析:過A′、B′分別作AB的垂線,垂足分別為E、D,由點A(1,-4),點B(5,-4),C為AB的中點可得到AB=4,C點坐標為(3,-4),再根據(jù)旋轉的性質得到∠A′CE=∠B′CD=30°,CA′=CB′=CB=2,根據(jù)含30度的直角三角形三邊的關系得到B′D=1,CD=CE=
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,則B′的坐標為(3+
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,-3),然后利用待定系數(shù)法確定線段A′B′的解析式,且自變量的范圍為3-
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≤x≤3+
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.
解答:如圖,
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過A′、B′分別作AB的垂線,垂足分別為E、D,
∵點A(1,-4),點B(5,-4),C為AB的中點,
∴AB=4,C點坐標為(3,-4),
∵線段AB繞中點C逆時針旋轉30°后,得到新的線段A′B′,
∴∠A′CE=∠B′CD=30°,CA′=CB′=CB=2,
∴B′D=1,CD=CE=
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,
∴B′的坐標為(3+
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,-3)
設線段A′B′的解析式為y=kx+b,
把C(3,-4)、B′(3+
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,-3)代入
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,解得
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,
∴線段A′B′的解析式為y=
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x-4-
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(3-
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≤x≤3+
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).
故答案為y=
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x-4-
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(3-
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≤x≤3+
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).
點評:本題考查了待定系數(shù)法求一次函數(shù)的解析式:設一次函數(shù)的解析式為y=kx+b(k、b為常數(shù),k≠0),然后把一次函數(shù)圖象的兩個點的坐標代入得到關于k、b的方程組,再解方程組得到k、b的值,從而確定一次函數(shù)的解析式.也考查了旋轉的性質.