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

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(C) 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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(C) 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 $m$. The angle between the lines is $60^o$,so $	an(60^o) = |\frac{m - m_2}{1 + m \cdot m_2}|$.$\sqrt{3} = |\frac{m - (-\sqrt{3})}{1 + m(-\sqrt{3})}| = |\frac{m + \sqrt{3}}{1 - \sqrt{3}m}|$.Squaring both sides: $3 = \frac{(m + \sqrt{3})^2}{(1 - \sqrt{3}m)^2} \Rightarrow 3(1 - 2\sqrt{3}m + 3m^2) = m^2 + 2\sqrt{3}m + 3$.$3 - 6\sqrt{3}m + 9m^2 = m^2 + 2\sqrt{3}m + 3 \Rightarrow 8m^2 - 8\sqrt{3}m = 0$.$8m(m - \sqrt{3}) = 0$,so $m = 0$ or $m = \sqrt{3}$.If $m = 0$,the line is $y + 2 = 0(x - 3) \Rightarrow y + 2 = 0$,which is parallel to the $x$-axis and does not intersect it (unless it is the $x$-axis itself,which it is not).If $m = \sqrt{3}$,the line is $y - (-2) = \sqrt{3}(x - 3) \Rightarrow y + 2 = \sqrt{3}x - 3\sqrt{3}$.Rearranging gives $y - \sqrt{3}x + 2 + 3\sqrt{3} = 0$.

$A$ straight line $L$ through the point $(3, -2)$ is inclined at an angle of $60^o$ 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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