直线方程y=kx(k=tana)
设任意点(x,y)经反射后变成(x',y')
有(y+y')/2=k(x+x')/2(中点在直线上)
(y'-y)/(x'-x)=-1/k(垂直）
y+y'=k(x+x')
y'-y=-(x'-x)/k
解得x'=(1-kk)x/(1+kk)+2ky/(kk+1),y'=2kx/(kk+1)+(kk-1)y/(kk+1)
即x'=xcos2a+ysin2a
y'=xsin2a-ycos2a