A ray of light is incident along a line which meets another line $$7x - y + 1 = 0$$ at the point $$(0, 1)$$. The ray is then reflected from this point along the line $$y + 2x = 1$$. Then the equation of the line of incidence of the ray of light is:

We have the mirror (reflecting surface) given by the straight-line equation $$7x - y + 1 = 0$$. Its normal vector is obtained directly from its coefficients, so we take $$\vec n = (7,\,-1)$$. A direction vector along the mirror (i.e. tangent to the surface) must be perpendicular to this normal. One convenient choice is obtained by swapping the components and changing one sign; hence we select $$\vec t = (1,\,7)$$ because $$\vec t \cdot \vec n = 1\cdot 7 + 7\cdot (-1)=0$$, confirming orthogonality.

The ray after reflection travels along the line $$y + 2x = 1 \; \Longrightarrow \; 2x + y - 1 = 0.$$ Writing it in slope form, $$y = 1 - 2x,$$ the slope is $$-2,$$ so a convenient direction vector for the reflected ray is $$\vec r = (1,\,-2).$$ Both the incident and reflected rays meet the mirror at the point $$P(0,1),$$ so all vectors can be regarded with origin at that point.

Law of reflection (in vector form): decompose the incident vector $$\vec i$$ into two perpendicular components, one parallel to the mirror ($$\vec t$$ direction) and one normal to the mirror ($$\vec n$$ direction). The parallel component remains unchanged on reflection, while the normal component reverses its sign. Symbolically, if

$$\vec i = \alpha\,\vec t + \beta\,\vec n,$$ then after reflection we get $$\vec r = \alpha\,\vec t - \beta\,\vec n.$$

Our task is therefore to (i) express the known reflected vector $$\vec r$$ in terms of $$\vec t$$ and $$\vec n,$$ find $$\alpha$$ and $$\beta,$$ and then (ii) construct the incident vector $$\vec i$$ by flipping the sign of the normal component.

Step (i): Decompose $$\vec r$$ Write $$\vec r = \alpha\,\vec t + \beta\,\vec n$$ or in component form $$(1,\,-2)=\alpha(1,7)+\beta(7,\,-1).$$ Equating components:

For the $$x$$-coordinate: $$1 = \alpha + 7\beta \qquad (1)$$ For the $$y$$-coordinate: $$-2 = 7\alpha - \beta \qquad (2)$$

From (1) we obtain $$\alpha = 1 - 7\beta.$$ Substituting in (2):

$$-2 = 7(1 - 7\beta) - \beta = 7 - 49\beta - \beta = 7 - 50\beta.$$ Thus $$-50\beta = -9 \;\Longrightarrow\; \beta = \dfrac{9}{50}.$$

Then $$\alpha = 1 - 7\left(\dfrac{9}{50}\right)=1-\dfrac{63}{50}= \dfrac{50 - 63}{50}= -\dfrac{13}{50}.$$

Hence the reflected vector can indeed be written as $$\vec r = -\dfrac{13}{50}\,\vec t \;+\; \dfrac{9}{50}\,\vec n.$$

Step (ii): Construct the incident vector $$\vec i$$ According to the reflection rule, just reverse the sign of the normal component:

$$\vec i = \alpha\,\vec t \;-\; \beta\,\vec n = -\dfrac{13}{50}\,\vec t \;-\; \dfrac{9}{50}\,\vec n.$$

Insert $$\vec t = (1,7)$$ and $$\vec n = (7,-1):$$

$$\vec i = -\dfrac{13}{50}(1,7)\;-\;\dfrac{9}{50}(7,-1) = \left(-\dfrac{13}{50},\,-\dfrac{91}{50}\right) + \left(-\dfrac{63}{50},\,\dfrac{9}{50}\right) = \left(-\dfrac{76}{50},\, -\dfrac{82}{50}\right).$$

Simplify the common factor $$\dfrac{1}{50}:$$

$$\vec i = \left(-\dfrac{38}{25},\, -\dfrac{41}{25}\right).$$

Removing the denominator (multiply by $$25$$) gives the direction vector $$\vec i = (-38,\,-41).$$ The negative sign merely indicates direction; any non-zero scalar multiple represents the same straight line. For convenience we take the positive multiple $$\vec d = (38,\,41).$$

Equation of the incident line
The required line must pass through $$P(0,1)$$ and have direction vector $$\vec d = (38,41).$$ Slope $$m$$ of this vector is $$m = \dfrac{41}{38}.$$ Using the point-slope form,

$$y - 1 = m(x - 0) \;\Longrightarrow\; y - 1 = \dfrac{41}{38}\,x.$$

Eliminate the fraction by multiplying by $$38$$:

$$38(y - 1) = 41x \;\Longrightarrow\; 38y - 38 = 41x.$$

Rearrange to standard form:

$$41x - 38y + 38 = 0.$$

This matches Option C among the given choices.

Hence, the correct answer is Option C.