A straight line L through the point (3, -2) i | vedclass

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(B) Let the slope of line $L$ be $m$. The slope of the given line $\sqrt{3}x + y = 1$ is $m_1 = -\sqrt{3}$.Given the angle between the lines is $60^{\

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

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(B) Let the slope of line $L$ be $m$. The slope of the given line $\sqrt{3}x + y = 1$ is $m_1 = -\sqrt{3}$.Given the angle between the lines is $60^{\circ}$,we have $	an 60^{\circ} = \left| \frac{m - m_1}{1 + m \cdot m_1} ight|$.$\sqrt{3} = \left| \frac{m - (-\sqrt{3})}{1 + m(-\sqrt{3})} ight| = \left| \frac{m + \sqrt{3}}{1 - \sqrt{3}m} ight|$.Case $1$: $\frac{m + \sqrt{3}}{1 - \sqrt{3}m} = \sqrt{3}$ $\Rightarrow m + \sqrt{3} = \sqrt{3} - 3m$ $\Rightarrow 4m = 0$ $\Rightarrow m = 0$.The equation of line $L$ is $y - (-2) = 0(x - 3) \Rightarrow y + 2 = 0$.Case $2$: $\frac{m + \sqrt{3}}{1 - \sqrt{3}m} = -\sqrt{3}$ $\Rightarrow m + \sqrt{3} = -\sqrt{3} + 3m$ $\Rightarrow 2m = 2\sqrt{3}$ $\Rightarrow m = \sqrt{3}$.The equation of line $L$ is $y - (-2) = \sqrt{3}(x - 3)$ $\Rightarrow y + 2 = \sqrt{3}x - 3\sqrt{3}$ $\Rightarrow y - \sqrt{3}x + 2 + 3\sqrt{3} = 0$.

$A$ straight line $L$ through the point $(3, -2)$ is inclined at an angle $60^{\circ}$ 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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