摘要: 解:(1)由直角三角形纸板的两直角边的长为1和2. 知两点的坐标分别为. 设直线所对应的函数关系式为.···························································· 2分 有解得 所以.直线所对应的函数关系式为.····················································· 4分 (2)①点到轴距离与线段的长总相等. 因为点的坐标为. 所以.直线所对应的函数关系式为. 又因为点在直线上. 所以可设点的坐标为. 过点作轴的垂线.设垂足为点.则有. 因为点在直线上.所以有.······················· 6分 因为纸板为平行移动.故有.即. 又.所以. 法一:故. 从而有. 得.. 所以. 又有.························································ 8分 所以.得.而. 从而总有.····································································································· 10分 法二:故.可得. 故. 所以. 故点坐标为. 设直线所对应的函数关系式为. 则有解得 所以.直线所对的函数关系式为.·············································· 8分 将点的坐标代入.可得.解得. 而.从而总有.························································· 10分 ②由①知.点的坐标为.点的坐标为. .····································································· 12分 当时.有最大值.最大值为. 取最大值时点的坐标为. 14分

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