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(B) Let the given line be $L_1: \sqrt{3}x + y - 9 = 0$. The slope of $L_1$ is $m_1 = -\sqrt{3}$.Let the slope of the required line $L$ be $m$. The ang

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$\sqrt{3}x - y - 3 - 5\sqrt{3} = 0$

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(B) Let the given line be $L_1: \sqrt{3}x + y - 9 = 0$. The slope of $L_1$ is $m_1 = -\sqrt{3}$.Let the slope of the required line $L$ be $m$. The angle between $L$ and $L_1$ is $60^{\circ}$.Using the formula $	an 	heta = \left| \frac{m - m_1}{1 + m \cdot m_1} ight|$,we have:$	an 60^{\circ} = \left| \frac{m - (-\sqrt{3})}{1 + m(-\sqrt{3})} ight|$ $\Rightarrow \sqrt{3} = \left| \frac{m + \sqrt{3}}{1 - \sqrt{3}m} ight|$.Case $1$: $\sqrt{3} = \frac{m + \sqrt{3}}{1 - \sqrt{3}m}$ $\Rightarrow \sqrt{3} - 3m = m + \sqrt{3}$ $\Rightarrow 4m = 0$ $\Rightarrow m = 0$.The equation is $y - (-3) = 0(x - 5) \Rightarrow y + 3 = 0$. This line is parallel to the $X$-axis and does not intersect it at a unique point (it is the $X$-axis itself if $y=0$,but here $y=-3$).Case $2$: $-\sqrt{3} = \frac{m + \sqrt{3}}{1 - \sqrt{3}m}$ $\Rightarrow -\sqrt{3} + 3m = m + \sqrt{3}$ $\Rightarrow 2m = 2\sqrt{3}$ $\Rightarrow m = \sqrt{3}$.The equation of line $L$ is $y - (-3) = \sqrt{3}(x - 5)$ $\Rightarrow y + 3 = \sqrt{3}x - 5\sqrt{3}$ $\Rightarrow \sqrt{3}x - y - 3 - 5\sqrt{3} = 0$.

If a straight line $L$ passing through the point $(5, -3)$ is inclined at an angle of $60^{\circ}$ to the line $\sqrt{3}x + y - 9 = 0$ and $L$ intersects the $X$-axis,then the equation of $L$ is

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