Question

A ray of light travelling along the line x+√3y=5 is incident on the x-axis and after reflection, it enter the other side of the x-axis by turning π\6 away from the x-axis . the equation of the line along which the reflected ray travels is ?

Answer 1

The equation of the reflected ray is:

$\boxed{x+\sqrt{3}y=5}$

Step-by-step explanation:

Equation of the line

$x+\sqrt{3}y=5$

Putting y = 0 in this equation

We get

$x=5$

The given line touches the x-axis at $(5,0)$

The refracted line makes angle from the x-axis $=-\frac{\pi}{6}$

The slope of the refracted line will be

$m=\tan(-\frac{\pi}{6})$

$\implies m=-\frac{1}{\sqrt{3}}$

Thus the equation of the refracted line is

$y-0=-\frac{1}{\sqrt{3}}(x-5)$

$\implies \sqrt{3}y=-x+5$

$\implies x+\sqrt{3}y=5$

Answer 2

• Answer:

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