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A line through the point $\mathrm{A}(2,0)$ which makes an angle of $30^{\circ}$ with the positive direction of $x$-axis is rotated about A in clockwise direction through an angle $15^{\circ}$. Then the equation of the straight line in the new position is
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Verified Answer
The correct answer is:
$(2-\sqrt{3}) x-y-4+2 \sqrt{3}=0$
Hints : Equation of line in new position :
$$
\begin{aligned}
& y-0=\tan 15^{\circ}(x-2) \\
& \Rightarrow y=\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)(x-2) \\
& \Rightarrow y=\frac{(\sqrt{3}-1)^2}{2}(x-2)
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow 2 y=(4-2 \sqrt{3})(x-2) \\
& \Rightarrow y=(2-\sqrt{3})(x-2) \\
& \Rightarrow(2-\sqrt{3}) x-y-4+2 \sqrt{3}=0
\end{aligned}
$$
$$
\begin{aligned}
& y-0=\tan 15^{\circ}(x-2) \\
& \Rightarrow y=\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)(x-2) \\
& \Rightarrow y=\frac{(\sqrt{3}-1)^2}{2}(x-2)
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow 2 y=(4-2 \sqrt{3})(x-2) \\
& \Rightarrow y=(2-\sqrt{3})(x-2) \\
& \Rightarrow(2-\sqrt{3}) x-y-4+2 \sqrt{3}=0
\end{aligned}
$$
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