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A ray of light travelling along the line $\mathrm{2 x-3 y+5=0}$ after striking a plane mirror lying along the line $\mathrm{x+y=2}$ gets reflected. Find the equation of the straight line containing the reflected ray.

Option: 1
$\mathrm{2 x-3 y+3=0}$

Option: 2
$\mathrm{3 x-2 y+3=0}$

Option: 3
$\mathrm{3 x+2 y-3=0}$

Option: 4
$\mathrm{2 y=3 x-2}$

Answers (1)

best_answer

The point of intersection of the lines $\mathrm{2 x-3 y+5=0}$ and $\mathrm{x+y=2}$ is $\mathrm{\left(\frac{1}{5}, \frac{9}{5}\right). }$

$\mathrm{\left(\frac{1}{5}, \frac{9}{5}\right). }$ is the point of incidence.

Slope m of the normal to the mirror (i.e. normal to the line $\mathrm{x+y=2 }$ ) is 1 .

Now the incident ray and reflected ray both are equally inclined to the normal and are on opposite side of it.
Slope of incident ray $\mathrm{m_I=\frac{2}{3}}$

Let the slope of the reflected ray be = $\mathrm{m_2}$

Then
$\mathrm{\frac{m_1-m}{1+m_1 m}=\frac{m-m_2}{1+m_2 m} }$

i.e $\mathrm{\frac{\frac{2}{3}-1}{1+\frac{2}{3} \times 1}=\frac{1-m_2}{1+m_2 \times 1} }$

$\mathrm{\therefore m_2=\frac{3}{2} }$ , $\mathrm{\therefore }$ the equation of the straight line containing the reflected ray is

$\mathrm{y-\frac{9}{5}=\frac{3}{2}\left(x-\frac{1}{5}\right) \quad \text { i.e. } \quad 3 x-2 y+3=0 }$

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