JEE Mains · Maths · STD 11 - 9. straight line
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
- A \(\frac{2}{(\sqrt{3}-1)}\)
- B \(\frac{2}{3+\sqrt{3}}\)
- C \(\frac{2}{3-\sqrt{3}}\)
- D \(\frac{\sqrt{3}}{2(\sqrt{3}+1)}\)
Answer & Solution
Correct Answer
(B) \(\frac{2}{3+\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
Slope of reflected ray \(=\tan 60^{\circ}=\sqrt{3}\) Line \(y=\frac{x}{\sqrt{3}}\) intersect \(y+x=1\) at \(\left(\frac{\sqrt{3}}{\sqrt{3}+1}, \frac{1}{\sqrt{3}+1}\right)\) Equation of reflected ray is \(y-\frac{1}{\sqrt{3}+1}=\sqrt{3}\left(x-\frac{\sqrt{3}}{\sqrt{3}+1}\right)\)…
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