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

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(B) The equation of a straight line passing through $(3,-2)$ with slope $m$ is given by $y+2=m(x-3)$ $(i)$.The given line is $\sqrt{3} x+y=1$,which ca

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

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(B) The equation of a straight line passing through $(3,-2)$ with slope $m$ is given by $y+2=m(x-3)$ $(i)$.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 $m_1=-\sqrt{3}$.The angle between the two lines is $60^{\circ}$. Using the formula $	an 	heta = \left| \frac{m-m_1}{1+m m_1} ight|$,we have $	an 60^{\circ} = \left| \frac{m-(-\sqrt{3})}{1+m(-\sqrt{3})} ight|$.$\sqrt{3} = \left| \frac{m+\sqrt{3}}{1-\sqrt{3} m} ight|$.Case $1$: $\sqrt{3} = \frac{m+\sqrt{3}}{1-\sqrt{3} m} \Rightarrow \sqrt{3} - 3m = m + \sqrt{3} \Rightarrow 4m = 0 \Rightarrow m = 0$.Substituting $m=0$ into $(i)$,we get $y+2=0(x-3) \Rightarrow y+2=0$. This line is parallel to the $X$-axis and does not intersect the $X$-axis (unless it is the $X$-axis itself,which it is not).Case $2$: $-\sqrt{3} = \frac{m+\sqrt{3}}{1-\sqrt{3} m} \Rightarrow -\sqrt{3} + 3m = m + \sqrt{3} \Rightarrow 2m = 2\sqrt{3} \Rightarrow m = \sqrt{3}$.Substituting $m=\sqrt{3}$ into $(i)$,we get $y+2=\sqrt{3}(x-3) \Rightarrow y+2=\sqrt{3}x-3\sqrt{3} \Rightarrow y-\sqrt{3}x+2+3\sqrt{3}=0$.

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