直线L过点a(3,2)且被L1：X-3Y+10=0,L2:2X-Y-8=0所截的线段恰以a为中点,求直线L的方程

设直线L的方程y-2=k(x-3)x-3y+10=0y1=(13k-2)/(3k-1)x1=(9k+4)/(3k-1)交点（x1,y1）y-2=k(x-3)2x-y-8=0x2=(3k-10)/(k-2)y2=(2k+4)/(2-k)恰以a为中点x1+x2=6y1+y2=4k=-2直线L的方程y=-2x+8