# A ray of light coming from the general eqn of the pt $\left ( 1,2\sqrt{3} \right )$ is incident at an angle  $30^{\circ}$ on the line x=1 at point A. The ray gets reflected on the line x=1 and meets x-axis at pt.B Then the line AB pass via ... Option: 1 Option: 2 Option: 3 Option: 4

$\\\tan 60^{\circ}=\frac{2 \sqrt{3}-\mathrm{k}}{2-1} \\ \\\sqrt{3}=2 \sqrt{3}-\mathrm{k} \\ \\\therefore \mathrm{k}=\sqrt{3}$

\begin{aligned} &\text { So point } \mathrm{A}(1, \sqrt{3})\\ &\text { Now slope of line } \mathrm{AB} \text { in } \mathrm{m}_{\mathrm{AB}}=\tan 120^{\circ} \end{aligned}

\begin{aligned} &\mathrm{mm}_{\mathrm{AB}}=-\sqrt{3}\\ &\text {Now equation of line } \mathrm{AB} \text { is } \end{aligned}

\begin{aligned} &y-\sqrt{3}=-\sqrt{3}(x-1)\\ &\sqrt{3} x+y=2 \sqrt{3}\\ &\text {Now satisfy options } \end{aligned}

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