设直线方程为L：y-3=k(x-2)
即：kx-y-2k+3=0
直线与点的距离d=|-k-(-2)-2k+3|/√（k^2+1）=|7k-4-2k+3|/√(k^2+1)
于是有[-k-(-2)-2k+3]^2=[7k-4-2k+3]^2
(3k-5)^2=(5k-1)^2
4k^2+5k-6=0
k=-2,或k=3/4
k=-2时,L：y=-2x+7
k=3/4时,L：y=3/4x+3/2