1. Given lines

• Reflecting mirror line: $7x-y+1=0 \quad \Rightarrow \quad y=7x+1$ Its slope is $m_1=7$

• Reflected ray line: $y+2x=1 \quad \Rightarrow \quad y=1-2x$ So its slope is $m_r=-2$

The point of reflection is given as $(0,1)$, and indeed both lines pass through it.

2. Use the reflection property

When a ray reflects from a line, the mirror line bisects the angle between the incident ray and reflected ray.

So if the incident line has slope $m$, then the angles made with the mirror line and reflected line satisfy the angle-bisector condition.

A convenient formula is: $\tan \theta=\left|\frac{m_a-m_b}{1+m_am_b}\right|$ for angle between two lines of slopes $m_a, m_b$.

Thus, $\left|\frac{m-7}{1+7m}\right|=\left|\frac{-2-7}{1+7(-2)}\right|$

Compute the right side: $\left|\frac{-9}{1-14}\right|=\left|\frac{-9}{-13}\right|=\frac{9}{13}$

So, $\frac{m-7}{1+7m}=\pm \frac{9}{13}$

3. Solve for possible slopes of incident line

Case 1:

$\frac{m-7}{1+7m}=\frac{9}{13}$ $13(m-7)=9(1+7m)$ $13m-91=9+63m$ $-100=50m$ $m=-2$

This is the same as the reflected ray, so this is not the incident ray.

Case 2:

$\frac{m-7}{1+7m}=-\frac{9}{13}$ $13(m-7)=-9(1+7m)$ $13m-91=-9-63m$ $76m=82$ $m=\frac{41}{38}$

Hence the incident ray has slope $m=\frac{41}{38}$

4. Equation of the incident line

It passes through $(0,1)$, so using point-slope form: $y-1=\frac{41}{38}(x-0)$ $38y-38=41x$ $41x-38y+38=0$

5. Match with options

This is exactly:

A: $41x-38y+38=0$

6. Comparison with stored correct answer

Stored correct answer: A

Derived answer: A

They agree.