A line L passes through the point (3, -2) and | vedclass

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(B) The given line is $\sqrt{3}x + y = 1$,which can be written as $y = -\sqrt{3}x + 1$. The slope of this line is $m_2 = -\sqrt{3}$.Let the slope of l

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$y - \sqrt{3}x + 2 + 3\sqrt{3} = 0$

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(B) The given line is $\sqrt{3}x + y = 1$,which can be written as $y = -\sqrt{3}x + 1$. The slope of this line is $m_2 = -\sqrt{3}$.Let the slope of line $L$ be $m_1$. The angle between the lines is $60^{\circ}$.Using the formula $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$,we have:$	an 60^{\circ} = \left| \frac{m_1 - (-\sqrt{3})}{1 + m_1(-\sqrt{3})} ight|$$\sqrt{3} = \left| \frac{m_1 + \sqrt{3}}{1 - \sqrt{3}m_1} ight|$This gives two cases:Case $1$: $\frac{m_1 + \sqrt{3}}{1 - \sqrt{3}m_1} = \sqrt{3}$ $\Rightarrow m_1 + \sqrt{3} = \sqrt{3} - 3m_1$ $\Rightarrow 4m_1 = 0$ $\Rightarrow m_1 = 0$.The equation of line $L$ with slope $0$ passing through $(3, -2)$ is $y - (-2) = 0(x - 3) \Rightarrow y + 2 = 0$.Case $2$: $\frac{m_1 + \sqrt{3}}{1 - \sqrt{3}m_1} = -\sqrt{3}$ $\Rightarrow m_1 + \sqrt{3} = -\sqrt{3} + 3m_1$ $\Rightarrow 2m_1 = 2\sqrt{3}$ $\Rightarrow m_1 = \sqrt{3}$.The equation of line $L$ with slope $\sqrt{3}$ passing through $(3, -2)$ is $y - (-2) = \sqrt{3}(x - 3)$ $\Rightarrow y + 2 = \sqrt{3}x - 3\sqrt{3}$ $\Rightarrow y - \sqrt{3}x + 2 + 3\sqrt{3} = 0$.

$A$ line $L$ passes through the point $(3, -2)$ and is inclined at an angle of $60^{\circ}$ to the line $\sqrt{3}x + y = 1$. If $L$ also intersects the $x$-axis,then the equation of $L$ is:

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