l:x-y+3=0,光线从A(1,2)射向x 上的B，从B反射到l上C，从C反射回A

设 B 点坐标是 (m, 0)

直线l: x-y+3 与x轴的交点是 (-3,0)
A 点坐标是 (1,2)
则必须有 -3 < m < 1

为叙述方便，定义 直线l 与 x轴 和y轴分别交于 D E。

设 ∠ABC = b

如果题目所叙述的光路能够实现，那么应有：
∠CBD = (180-b)/2 = 90 - b/2

直线l的斜率为1。与x轴正方向的夹角为 45 度
∠DCB = 180 - ∠CDB - ∠CBD
= 180 - 45 - (90 - b/2)
= 45 + b/2
∠ECA = ∠DCB = 45 + b/2

∠BCA = 180 - ∠ECA - ∠DCB = 180 - 2(45+b/2) = 90 - b
∠BAC = 180 - ∠ABC - ∠BCA = 180 - b - (90 - b) = 90

即，如题目中所叙述的光路能够实现，则
AC⊥AB

AB 直线的斜率
k1 = (2 -0)/(1-m) = 2/(1-m)

BC 直线的斜率
k2 = -k1 = 2/(m-1)

BC 直线的方程：
y - 0 = k2(x-m)
y = [2/(m-1)] (x - m)

联立 BC 直线方程 与 l的方程，求 C 点坐标
y = [2/(m-1)] (x - m)
x-y+3 = 0

x + 3 = [2/(m-1)](x-m)
(m-1)(x+3) = 2(x-m)
(m-1)x + 3m -3 = 2x - 2m
(m-3)x = 3 - 5m
x = (3-5m)/(m-3)
y = x + 3 = (3-5m)/(m-3) + 3 = (3-5m + 3m - 9)/(m-3) = (-