设直线L的方程为
y-1=k(x-3)
与直线L1的交点坐标为：[3k/(k+1),(1-2k)/(k+1)]
与直线L2的交点坐标为：[(3k+5)/(k+1),(3k+1)/(k+1)]
两点之间的距离为d
d=√[(3k+1)/(k+1)-(1-2k)/(k+1)]^2+[(3k+5)/(k+1)-3k/(k+1)]^2
=5/(k+1)√k^2+1
=5
求得k=0
L:y=1