求出A关于直线的对称点C(a,b)
则AC垂直直线,AC中点在直线上
x+y+1=0斜率=-1
则AC斜率=1
(b-3)/(a-2)=1
b-3=a-2
a-b=-1
AC中点[(a+2)/2,(b+3)/2]在直线上
(a+2)/2+(b+3)/2+1=0
a+2+b+3+2=0
a+b=-7
a-b=-1
a=-4,b=-3
C(-4,-3)反射光线是BC
(y+3)/(1+3)=(x+2)/(1+2)
4x-3y-1=0
反射光线和x+y+1=0交点
D(-2/7,-5/7)
入射光线AD
13x-8y-2=0