2x+y-1=0 →l1
x+y+1=0 →l2
兩線交點(diǎn)(2,-3)
取2x+y-1=0上一點(diǎn)P(0,1)
設(shè)P'(x',y')為P關(guān)于l2的對(duì)稱點(diǎn),則
PP'與l2垂直→m(PP')*m(l2)=-1
→m(PP')=-1/-1=1
PP'=l3→P,m(PP')→y=x+1
l2與l3的交點(diǎn):(-1,0)為PP'中點(diǎn)
(0+x'/2,1+y'/2)=(-1,0),所以P'=(-2,-1)
過點(diǎn)(2,3)和P'做直線,則答案為
x+2y-1=0
兩線關(guān)于某條線對(duì)稱,真要說的話可以用兩線夾角判斷..相對(duì)于對(duì)稱線夾角一樣大