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(B) The line $L$ is at a distance $d=4$ from the origin. Let the perpendicular from the origin to $L$ make an angle $\alpha$ with the positive $x$-axi

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$(\sqrt{3}-1) x+(\sqrt{3}+1) y=8 \sqrt{2}$

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(B) The line $L$ is at a distance $d=4$ from the origin. Let the perpendicular from the origin to $L$ make an angle $\alpha$ with the positive $x$-axis. The equation of the line $L$ is $x \cos \alpha + y \sin \alpha = 4$.Given that the perpendicular makes an angle of $60^{\circ}$ with the line $x+y=0$. The line $x+y=0$ makes an angle of $135^{\circ}$ with the positive $x$-axis.Since the line $L$ makes positive intercepts,the perpendicular must lie in the first quadrant,so $0 The angle between the perpendicular and $x+y=0$ is $135^{\circ} - \alpha = 60^{\circ}$,which gives $\alpha = 75^{\circ}$.Now,$\cos 75^{\circ} = \cos(45^{\circ}+30^{\circ}) = \cos 45^{\circ} \cos 30^{\circ} - \sin 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}-1}{2\sqrt{2}}$.$\sin 75^{\circ} = \sin(45^{\circ}+30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$.Substituting these into the normal form equation: $x \left( \frac{\sqrt{3}-1}{2\sqrt{2}} \right) + y \left( \frac{\sqrt{3}+1}{2\sqrt{2}} \right) = 4$.Multiplying by $2\sqrt{2}$,we get $(\sqrt{3}-1)x + (\sqrt{3}+1)y = 8\sqrt{2}$.

$A$ straight line $L$ at a distance of $4$ units from the origin makes positive intercepts on the coordinate axes,and the perpendicular drawn from the origin to this line makes an angle of $60^{\circ}$ with the line $x+y=0$. Then the equation of the line $L$ is

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