A line passing through the point P(a, 0) make | vedclass

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(A) Let the initial line be $PQ$ with slope $	an \alpha$. After rotating clockwise by $\frac{\alpha}{2}$,the new line $PR$ has an angle of inclinatio

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$4$

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(A) Let the initial line be $PQ$ with slope $	an \alpha$. After rotating clockwise by $\frac{\alpha}{2}$,the new line $PR$ has an angle of inclination $\alpha - \frac{\alpha}{2} = \frac{\alpha}{2}$.Given the slope of the new line $PR$ is $2-\sqrt{3}$,we have $	an(\frac{\alpha}{2}) = 2-\sqrt{3} = 	an 15^{\circ}$.Thus,$\frac{\alpha}{2} = 15^{\circ}$,which implies $\alpha = 30^{\circ}$.The equation of line $PR$ passing through $P(a, 0)$ with slope $m = 2-\sqrt{3}$ is $y - 0 = (2-\sqrt{3})(x - a)$,which simplifies to $(2-\sqrt{3})x - y - a(2-\sqrt{3}) = 0$.The perpendicular distance from the origin $(0, 0)$ to this line is given as $\frac{1}{\sqrt{2}}$.Using the distance formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$,we get $\frac{|-a(2-\sqrt{3})|}{\sqrt{(2-\sqrt{3})^2 + (-1)^2}} = \frac{1}{\sqrt{2}}$.Simplifying the denominator: $\sqrt{4 + 3 - 4\sqrt{3} + 1} = \sqrt{8 - 4\sqrt{3}} = \sqrt{2(4 - 2\sqrt{3})} = \sqrt{2}(\sqrt{3}-1)$.So,$\frac{|a|(2-\sqrt{3})}{\sqrt{2}(\sqrt{3}-1)} = \frac{1}{\sqrt{2}}$.Since $2-\sqrt{3} = \frac{(\sqrt{3}-1)^2}{2}$ is not quite right,let's use $\sqrt{8-4\sqrt{3}} = \sqrt{2}\sqrt{4-2\sqrt{3}} = \sqrt{2}(\sqrt{3}-1)$.$|a| = \frac{\sqrt{2}(\sqrt{3}-1)}{\sqrt{2}(2-\sqrt{3})} = \frac{\sqrt{3}-1}{2-\sqrt{3}} 	imes \frac{2+\sqrt{3}}{2+\sqrt{3}} = 2\sqrt{3} + 3 - 2 - \sqrt{3} = \sqrt{3}+1$.$a^2 = (\sqrt{3}+1)^2 = 3+1+2\sqrt{3} = 4+2\sqrt{3}$.Now,calculate $3a^2 	an^2 \alpha - 2\sqrt{3} = 3(4+2\sqrt{3}) 	an^2 30^{\circ} - 2\sqrt{3} = 3(4+2\sqrt{3}) \cdot \frac{1}{3} - 2\sqrt{3} = 4+2\sqrt{3} - 2\sqrt{3} = 4$.

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