点（0,-1）是直线在y轴截距
所以设直线为y=kx-1即kx-y-1=0
点（1,-3）到直线距离=3√2/2
有｜k+3-1｜/√(1+k²)=3√2/2
｜k+2｜/√(1+k²)=3/√2
(k²+4k+4)/(k²+1)=9/2
2k²+8k+8=9k²+9
7k²-8k+1=0
(7k-1)(k-1)=0
k=1/7或k=1
所以直线为1/7x-y-1=0即x-7y-7=0或x-y-1=0