Key Concepts and Formulas

• Law of Reflection: The angle of incidence equals the angle of reflection.
• Slope-intercept form of a line: $y = mx + c$, where $m$ is the slope.
• Tangent of the angle between two lines: If $\theta$ is the angle between two lines with slopes $m_1$ and $m_2$, then $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.

Step-by-Step Solution

Step 1: Find the slopes of the reflecting line and the reflected ray.

We are given the equations of the reflecting line (mirror) and the reflected ray. We need to find their slopes by rewriting them in slope-intercept form.

• Reflecting line: $7x - y + 1 = 0$. Rewrite as $y = 7x + 1$. Therefore, the slope of the reflecting line is $m_M = 7$.

• Reflected ray: $y + 2x = 1$. Rewrite as $y = -2x + 1$. Therefore, the slope of the reflected ray is $m_R = -2$.

Step 2: Apply the Law of Reflection using the tangent formula.

Let $m_I$ be the slope of the incident ray. According to the Law of Reflection, the angle between the incident ray and the reflecting line is equal to the angle between the reflected ray and the reflecting line. Therefore, the tangents of these angles are equal in magnitude.

$\left| \frac{m_I - m_M}{1 + m_I m_M} \right| = \left| \frac{m_R - m_M}{1 + m_R m_M} \right|$ Substitute $m_M = 7$ and $m_R = -2$: $\left| \frac{m_I - 7}{1 + 7m_I} \right| = \left| \frac{-2 - 7}{1 + (-2)(7)} \right|$ $\left| \frac{m_I - 7}{1 + 7m_I} \right| = \left| \frac{-9}{1 - 14} \right|$ $\left| \frac{m_I - 7}{1 + 7m_I} \right| = \left| \frac{-9}{-13} \right| = \frac{9}{13}$

Step 3: Solve for $m_I$.

The absolute value equation gives us two cases:

• Case 1: $\frac{m_I - 7}{1 + 7m_I} = \frac{9}{13}$ Cross-multiply: $13(m_I - 7) = 9(1 + 7m_I)$ $13m_I - 91 = 9 + 63m_I$ $-50m_I = 100$ $m_I = -2$

• Case 2: $\frac{m_I - 7}{1 + 7m_I} = -\frac{9}{13}$ Cross-multiply: $13(m_I - 7) = -9(1 + 7m_I)$ $13m_I - 91 = -9 - 63m_I$ $76m_I = 82$ $m_I = \frac{82}{76} = \frac{41}{38}$

Step 4: Determine the correct slope.

We have two possible slopes for the incident ray: $m_I = -2$ and $m_I = \frac{41}{38}$. Since $m_I = -2$ is the slope of the reflected ray, the slope of the incident ray must be $m_I = \frac{41}{38}$. The case where $m_I = m_R$ corresponds to the trivial case where the "incident" ray is actually the reflected ray.

Step 5: Find the equation of the incident ray.

We know that the incident ray passes through the point $(0, 1)$ and has a slope of $m_I = \frac{41}{38}$. Using the point-slope form of a line, $y - y_1 = m(x - x_1)$: $y - 1 = \frac{41}{38}(x - 0)$ $y - 1 = \frac{41}{38}x$ $38(y - 1) = 41x$ $38y - 38 = 41x$ $41x - 38y + 38 = 0$

Common Mistakes & Tips

• Forgetting the absolute value: The tangent formula gives the acute angle. Remember to consider both positive and negative cases when removing the absolute value.
• Incorrectly identifying the slopes: Ensure you rewrite the given equations in the slope-intercept form ($y = mx + c$) to correctly identify the slopes.
• Trivial Solution: One solution will always be the slope of the reflected ray. Discard that solution.

Summary

We used the Law of Reflection and the formula for the tangent of the angle between two lines to find the slope of the incident ray. By equating the tangents of the angles between the reflecting line and the incident and reflected rays, we found two possible slopes. We discarded the trivial solution (the slope of the reflected ray) and used the remaining slope and the point of incidence to determine the equation of the incident ray. The equation of the incident ray is $41x - 38y + 38 = 0$.

Final Answer

The final answer is $\boxed{41x - 38y + 38 = 0}$, which corresponds to option (A).