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(C) Let the slope of the incident ray be $m$. The mirror line is $7x - y + 1 = 0$,which has a slope $m_1 = 7$. The reflected ray is $y + 2x = 1$,which

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$41x - 38y + 38 = 0$

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(C) Let the slope of the incident ray be $m$. The mirror line is $7x - y + 1 = 0$,which has a slope $m_1 = 7$. The reflected ray is $y + 2x = 1$,which has a slope $m_2 = -2$. The point of incidence is $(0, 1)$.Since the angle of incidence equals the angle of reflection,the angle between the incident ray and the mirror is equal to the angle between the reflected ray and the mirror.Using the formula $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$,we have:$\left| \frac{m - 7}{1 + 7m} ight| = \left| \frac{7 - (-2)}{1 + 7(-2)} ight| = \left| \frac{9}{1 - 14} ight| = \left| \frac{9}{-13} ight| = \frac{9}{13}$.$\frac{m - 7}{1 + 7m} = \frac{9}{13}$ or $\frac{m - 7}{1 + 7m} = -\frac{9}{13}$.Case $1$: $13(m - 7) = 9(1 + 7m)$ $\Rightarrow 13m - 91 = 9 + 63m$ $\Rightarrow -50m = 100$ $\Rightarrow m = -2$. This is the slope of the reflected ray.Case $2$: $13(m - 7) = -9(1 + 7m)$ $\Rightarrow 13m - 91 = -9 - 63m$ $\Rightarrow 76m = 82$ $\Rightarrow m = \frac{41}{38}$.The equation of the incident line passing through $(0, 1)$ with slope $m = \frac{41}{38}$ is:$y - 1 = \frac{41}{38}(x - 0)$ $\Rightarrow 38y - 38 = 41x$ $\Rightarrow 41x - 38y + 38 = 0$.

$A$ ray of light incident along a line meets another line $7x - y + 1 = 0$ at the point $(0, 1)$ and 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

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