### Straight Lines Question 5

#### Question 5 - 29 January - Shift 1

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

(1) $\frac{2}{(\sqrt{3}-1)}$

(2) $\frac{2}{3+\sqrt{3}}$

(3) $\frac{2}{3-\sqrt{3}}$

(4) $\frac{\sqrt{3}}{2(\sqrt{3}+1)}$

## Show Answer

#### Answer: (2)

#### Solution:

#### Formula: Slope Formula

Slope of reflected ray $=\tan 60^{\circ}=\sqrt{ } 3$

Line $y=\frac{x}{\sqrt{3}}$ intersect $y+x=1$ at $(\frac{\sqrt{3}}{\sqrt{3}+1}, \frac{1}{\sqrt{3}+1})$

Equation of reflected ray is

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

Put $y=0 \Rightarrow x=\frac{2}{3+\sqrt{3}}$

Therefore, option (2) is correct.