A ray of light is incident along a line which | vedclass

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(C) Let the slope of the incident ray be $m$. The line $7x - y + 1 = 0$ has a slope $m_1 = 7$. The reflected ray $y + 2x = 1$ has a slope $m_2 = -2$.S

Vedclass · Accepted Answer

$41x - 38y + 38 = 0$

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(C) Let the slope of the incident ray be $m$. The line $7x - y + 1 = 0$ has a slope $m_1 = 7$. The reflected ray $y + 2x = 1$ has a slope $m_2 = -2$.Since the angle of incidence equals the angle of reflection,the angle between the incident ray and the mirror line equals the angle between the reflected ray and the mirror line.Using the formula for the angle between two lines $	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}$.Case $1$: $\frac{m - 7}{1 + 7m} = \frac{9}{13}$ $\Rightarrow 13m - 91 = 9 + 63m$ $\Rightarrow -50m = 100$ $\Rightarrow m = -2$. The line equation is $y - 1 = -2(x - 0) \Rightarrow 2x + y - 1 = 0$ (This is the reflected ray itself).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 line 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,$y + 2x = 1$. Then the equation of the line of incidence of the ray of light is

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