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(C) Let the equation of the incident ray be $y - 1 = m(x - 0)$,which is $mx - y + 1 = 0$. The angle between the incident ray and the mirror line $7x -

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$41x - 38y + 38 = 0$

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(C) Let the equation of the incident ray be $y - 1 = m(x - 0)$,which is $mx - y + 1 = 0$. The angle between the incident ray and the mirror line $7x - y + 1 = 0$ is equal to the angle between the reflected ray $2x + y - 1 = 0$ and the mirror line. The slope of the incident ray is $m_1 = m$,the slope of the mirror is $m_2 = 7$,and the slope of the reflected ray is $m_3 = -2$. Using the formula $	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}| = |\frac{m_2 - m_3}{1 + m_2 m_3}|$: $|\frac{m - 7}{1 + 7m}| = |\frac{7 - (-2)}{1 + 7(-2)}| = |\frac{9}{1 - 14}| = |\frac{9}{-13}| = \frac{9}{13}$. Case $1$: $\frac{m - 7}{1 + 7m} = \frac{9}{13}$ $\Rightarrow 13m - 91 = 9 + 63m$ $\Rightarrow -50m = 100$ $\Rightarrow m = -2$ (This is the reflected ray). Case $2$: $\frac{m - 7}{1 + 7m} = -\frac{9}{13}$ $\Rightarrow 13m - 91 = -9 - 63m$ $\Rightarrow 76m = 82$ $\Rightarrow m = \frac{41}{38}$. The equation of the incident ray is $y - 1 = \frac{41}{38}(x - 0)$ $\Rightarrow 38y - 38 = 41x$ $\Rightarrow 41x - 38y + 38 = 0$.

$A$ 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 $y + 2x = 1$. Find the equation of the line of incidence of the ray of light.

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