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  • A light ray emits from the origin making an angle  with the positive <img alt="\mathrm{x}" src="https://e

A light ray emits from the origin making an angle 30^{0} with the positive \mathrm{x}-axis. After getting reflected by the line \mathrm{x+y = 1,} if this ray intersects \mathrm{x-}axis at \mathrm{Q}, then the abscissa of \mathrm{Q} is

Option: 1

\frac{\sqrt{3}}{2\left ( \sqrt{3+1} \right )}


Option: 2

\frac{2}{3+\sqrt{3}}


Option: 3

\frac{2}{\left ( \sqrt{3}-1 \right )}


Option: 4

\frac{2}{ 3-\sqrt{3} }


Answers (1)

best_answer

Equation of ray is

\mathrm{y=\frac{1}{\sqrt3}x} ..............(1)

Image of \mathrm{0(0, 0)} in the line \mathrm{x + y = 1 }is lies on reflected ray

\mathrm{\frac{x-0}{1}=\frac{y-0}{1}=-2\frac{\left ( 0+0-1 \right )}{2}}

\Rightarrow \mathrm{M\left ( 1,1 \right )}

\therefore Point of Intersection of lines \mathrm{y=\frac{x}{\sqrt{3}}\: and\: x+y=1\: is\: p\left ( x,y \right )}

\therefore \mathrm{p}\left(\frac{3-\sqrt{3}}{2}, \frac{\sqrt{3}-1}{2}\right)
Now Reflected Ray is same as line passing through PM.
\therefore \text { Slope of } \mathrm{PM}=\frac{\frac{\sqrt{3}-1}{2}-1}{\frac{3-\sqrt{3}}{2}-1}=\frac{\sqrt{3}-3}{1-\sqrt{3}}=\sqrt{3}
Equation of \mathrm{PM} whose slope is \sqrt{3} and passing through \mathrm{M}(1,1).
\begin{aligned} & \mathrm{y}-1=\sqrt{3}(\mathrm{x}-1) \\ \end{aligned}

\begin{aligned} & \mathrm{y}=\sqrt{3} \mathrm{x}+(-\sqrt{3}+1) \\ \end{aligned}

\begin{aligned} & \because \text { ray, Intersects } \mathrm{x} \text {-axis at } \alpha(\mathrm{x}, 0) \\ \end{aligned}

\begin{aligned} & \therefore \mathrm{y}=0 \\ \end{aligned}

\begin{aligned} & \Rightarrow \sqrt{3} \mathrm{x}=-1(-\sqrt{3}+1) \Rightarrow \sqrt{3} \mathrm{x}=\sqrt{3}-1 \\ \end{aligned}

\begin{aligned} & \Rightarrow \mathrm{x}=1-\frac{1}{\sqrt{3}} \\ \end{aligned}

\begin{aligned} & \mathrm{x}=\frac{\sqrt{3}-1}{\sqrt{3}} \times \frac{\sqrt{3}+1}{(\sqrt{3}+1)}=\frac{2}{3+\sqrt{3}} \end{aligned}

\therefore$ abscissa of $\alpha$ is $\frac{2}{3+\sqrt{3}}

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