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A straight line L through the point (3, - 2) is inclined at an angle 60° to the line $\sqrt{3} x+y=1$ . If L also intersects the X-axis, there the equation of L is

Answer:

Explanation:

A straight lire passing through P and making an angle of  $\alpha=60^{0}$ is given by

 $\frac{y-y_{1}}{x-x_{1}}= \tan ( \theta \pm \alpha)$

 where  $\sqrt{3}x+y=1$

 $\Rightarrow$  $y=-\sqrt{3}x+1$

 Then , $\tan \theta= -\sqrt{3}$

 $\Rightarrow$  $\frac{y+2}{x-3}=\frac{\tan \theta \pm \tan \alpha}{ 1 \pm \tan \theta \tan \alpha}$

 $\Rightarrow$   $\frac{y+2}{y-3}=\frac{-\sqrt{3}+\sqrt{3}}{1-(-\sqrt{3})(\sqrt{3})}$

 and   $\frac{y+2}{x-3}= \frac{-\sqrt{3}-\sqrt{3}}{1+(-\sqrt{3})(\sqrt{3})}$

 $\Rightarrow$   y+2=0

 and      $\frac{y+2}{x-3}=\frac{-2 \sqrt{3}}{1-3}= \sqrt{3}$

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

 Neglecting , y+2=0  as it does not interest Y-axis