1.根据光的反射的性质，可知：入射线L1与2X-Y-7=0的夹角等于出射线L2与2X-Y-7=0的夹角设L1与L2在2X-Y-7=0上的交点为M(x,y)k1=(y-4)/(x+2), k2=(y-8)/(x-5), 直线2X-Y-7=0的斜率k=2∴(k-k1)/(1+k*k1)=(k2-k)/(1+k*k2).............(1)y=2x-7........................................(2)由(1)(2)解得:x=25/4, y=11/2由A(-2,4),B(5,8),M(25/4,11/2)可得:L1:2x-11y+48=0L2:2x+y-18=02.首先求出2X-Y-2=0与X-Y-3=0的交点M(-1,-4)然后求出M,P连线的斜率k1=(0+4)/(3+1)=1∵所截得线段恰好被点P平分∴L的斜率k2=-1/k1=-1∴L:x+y-3=03.可设M(t,t), N(t+2,t+2)根据A,B,M,N可求得AM:(t-5)(x-2)=(t-2)(y-5)BN:(t+1)(x+2)=(t+4)(y-1)∵直线AM与BN的交点正好落在Y轴上令x=0,消去y,得t=-1故M(-1,-1),N(1,1),交点(0,1)
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