题目

在同一平面内有2011条直线a1,a2,a3...,a2011,a1⊥a2,a2//a3,a3⊥a4,a4//a5,那么a1与a100的位置关系该如何

提问时间：2020-08-11

答案

a1⊥a2,a2‖a3→→a1⊥a3a1⊥a3,a3⊥a4→→a1‖a4a1‖a4,a4‖a5→→a1‖a5a1‖a5,a5⊥a6→→a1⊥a6a1⊥a6,a6‖a7→→a1⊥a7a1⊥a7,a7⊥a8→→a1‖a8a1‖a8,a8‖a9→→a1‖a9..看到规律了吧4个循环一次所以2011/4=502...