# A ray of light is sent along the line x+y=1 after being reflected from the line y_x=1 it is again reflected from the line y=0 then the equation of the line representing the ray after second reflection may be given as?

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Jane Share
Feb 9, 2018

Here, is a diagram of the whole thing happening,

Light after coming by the pathway of $y - x = 1$ or, $y = 1 + x$ falls on $Y$ axis at point $\left(0 , 1\right)$ (point of refelection is found as the solution of the given two equations of straight line,as that will be their point of meeting) and then gets refleted via pathway $x + y = 1$ or, $y = 1 - x$ then falls on $Y = 0$ i.e $X$ axis,and gets reflected as follows,

Now, clearly, slope of the straight line $y - x = 1$ is 45 degrees, and the reflected ray from the $X$ axis will be parallel to that.

So,let the equation of the reflected pathway is $y - x - c = 0$

And, that of the straight line,to which it will be parallel is $y = 1 + x$ or, $y - x - 1 = 0$

Now,if the distance between them is $d$ , we know, $d = \frac{| c 1 + c 2 |}{\sqrt{{a}^{2} + {b}^{2}}}$ for two straight lines $a x + b y + c 1 = 0$ & $a x + b y + c 2 = 0$

here, $a = 1 , b = 1$ and $d = \sqrt{{1}^{2} + {1}^{2}} = \sqrt{2}$

$d = \frac{| 1 - c |}{\sqrt{2}}$

So, $c = - 1$

So,the equation of straight line,alomg which the reflected ray will go, will be $y - x + 1 = 0$ or, $y = x - 1$

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