反比例函数与勾股定理结合(只能用初二知识解答:已知,如图17,直线Y=-2x+4交X轴于A,交Y轴于B,反比例函数y=x分之2与直线AB交于点C.(1)点C是线段AB的中点吗?为什么? (2)过C作CM⊥AN分别交OA于
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![反比例函数与勾股定理结合(只能用初二知识解答:已知,如图17,直线Y=-2x+4交X轴于A,交Y轴于B,反比例函数y=x分之2与直线AB交于点C.(1)点C是线段AB的中点吗?为什么? (2)过C作CM⊥AN分别交OA于](/uploads/image/z/1084606-70-6.jpg?t=%E5%8F%8D%E6%AF%94%E4%BE%8B%E5%87%BD%E6%95%B0%E4%B8%8E%E5%8B%BE%E8%82%A1%E5%AE%9A%E7%90%86%E7%BB%93%E5%90%88%EF%BC%88%E5%8F%AA%E8%83%BD%E7%94%A8%E5%88%9D%E4%BA%8C%E7%9F%A5%E8%AF%86%E8%A7%A3%E7%AD%94%EF%BC%9A%E5%B7%B2%E7%9F%A5%2C%E5%A6%82%E5%9B%BE17%2C%E7%9B%B4%E7%BA%BFY%3D-2x%2B4%E4%BA%A4X%E8%BD%B4%E4%BA%8EA%2C%E4%BA%A4Y%E8%BD%B4%E4%BA%8EB%2C%E5%8F%8D%E6%AF%94%E4%BE%8B%E5%87%BD%E6%95%B0y%3Dx%E5%88%86%E4%B9%8B2%E4%B8%8E%E7%9B%B4%E7%BA%BFAB%E4%BA%A4%E4%BA%8E%E7%82%B9C.%EF%BC%881%EF%BC%89%E7%82%B9C%E6%98%AF%E7%BA%BF%E6%AE%B5AB%E7%9A%84%E4%B8%AD%E7%82%B9%E5%90%97%3F%E4%B8%BA%E4%BB%80%E4%B9%88%3F+%EF%BC%882%EF%BC%89%E8%BF%87C%E4%BD%9CCM%E2%8A%A5AN%E5%88%86%E5%88%AB%E4%BA%A4OA%E4%BA%8E)
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