26．(1)解:由得点坐标为由得点坐标为 ∴··················································································· 由解得∴点的坐标为···································· ∴··························································· (2)解:∵点在上且 ∴点坐标为······················································································ 又∵点在上且 ∴点坐标为······················································································ ∴··········································································· (3)解法一:当时.如图1.矩形与重叠部分为五边形(时.为四边形)．过作于.则 ∴即∴ ∴ 即··································································· 29．问题解决如图(1).将正方形纸片折叠.使点落在边上一点(不与点.重合).压平后得到折痕．当时.求的值．类比归纳在图(1)中.若则的值等于 ,若则的值等于 ,若(为整数).则的值等于．(用含的式子表示) 联系拓广如图(2).将矩形纸片折叠.使点落在边上一点(不与点重合).压平后得到折痕设则的值等于．(用含的式子表示) 【查看更多】

题目列表(包括答案和解析)

解：（1）OA=1，OC=2

则A点坐标为（0，1），C点坐标为（2，0）

设直线AC的解析式为y=kx+b