The equations of the lines passing through th | vedclass

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(B) The given line is $x + y\sqrt{3} + 3\sqrt{3} = 0$,which can be written as $y = -\frac{1}{\sqrt{3}}x - 3$.The slope of this line is $m_1 = -\frac{1

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$x = 0, x - y\sqrt{3} = 0$

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(B) The given line is $x + y\sqrt{3} + 3\sqrt{3} = 0$,which can be written as $y = -\frac{1}{\sqrt{3}}x - 3$.The slope of this line is $m_1 = -\frac{1}{\sqrt{3}}$,so the angle it makes with the positive $x$-axis is $	heta_1 = 150^{\circ}$.Let the slope of the required lines passing through the origin be $m$. The angle between these lines and the given line is $60^{\circ}$.Using the formula $	an 	heta = |\frac{m - m_1}{1 + m m_1}|$,we have $	an 60^{\circ} = |\frac{m - (-1/\sqrt{3})}{1 + m(-1/\sqrt{3})}|$.$\sqrt{3} = |\frac{m\sqrt{3} + 1}{\sqrt{3} - m}|$.Case $1$: $\sqrt{3}(\sqrt{3} - m) = m\sqrt{3} + 1$ $\Rightarrow 3 - m\sqrt{3} = m\sqrt{3} + 1$ $\Rightarrow 2m\sqrt{3} = 2$ $\Rightarrow m = \frac{1}{\sqrt{3}}$. The line is $y = \frac{1}{\sqrt{3}}x$ or $x - y\sqrt{3} = 0$.Case $2$: $\sqrt{3}(\sqrt{3} - m) = -(m\sqrt{3} + 1)$ $\Rightarrow 3 - m\sqrt{3} = -m\sqrt{3} - 1$ $\Rightarrow 3 = -1$,which is impossible,implying the line is vertical $(m 	o \infty)$. The line is $x = 0$.Thus,the lines are $x = 0$ and $x - y\sqrt{3} = 0$.

The equations of the lines passing through the origin and making an angle of $60^{\circ}$ with the line $x + y\sqrt{3} + 3\sqrt{3} = 0$ are

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