JEE MAIN - Mathematics (2023 - 29th January Morning Shift - No. 9)
A light ray emits from the origin making an angle 30$$^\circ$$ with the positive $$x$$-axis. After getting reflected by the line $$x+y=1$$, if this ray intersects $$x$$-axis at Q, then the abscissa of Q is :
$${2 \over {\left( {\sqrt 3 - 1} \right)}}$$
$${2 \over {3 - \sqrt 3 }}$$
$${{\sqrt 3 } \over {2\left( {\sqrt 3 + 1} \right)}}$$
$${2 \over {3 + \sqrt 3 }}$$
Explanation
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$\because $ OP reflected by $x+y=1$.
So, image of $Q$ lies on $y=\frac{x}{\sqrt{3}}$
$$ \begin{aligned} & \therefore \quad \frac{x-h}{1}=\frac{y}{1}=\frac{-2(h-1)}{2} \\\\ & \therefore \quad x=1, y=1-h \end{aligned} $$
It lies on $y=\frac{x}{\sqrt{3}}$
$$ \begin{aligned} & \therefore \quad 1-h=\frac{1}{\sqrt{3}} \\\\ & \therefore \quad h=1-\frac{1}{\sqrt{3}}=\frac{\sqrt{3}-1}{\sqrt{3}}=\frac{2}{3+\sqrt{3}} \end{aligned} $$
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