Straight Line and Pair of Straight Lines 1 Question 23

23. 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

(2011)

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

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

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

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

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Solution:

  1. A straight line passing through $P$ and making an angle of $\alpha=60^{\circ}$, is given by

$$ \frac{y-y _1}{x-x _1}=\tan (\theta \pm \alpha) $$

$$ \Rightarrow \quad \sqrt{3} x+y=1 $$

$\Rightarrow y=-\sqrt{3} x+1$, then $\tan \theta=-\sqrt{3}$

$$ \begin{aligned} \frac{y+2}{x-3} & =\frac{\tan \theta \pm \tan \alpha}{1 \mp \tan \theta \tan \alpha} \\ \frac{y+2}{x-3} & =\frac{-\sqrt{3}+\sqrt{3}}{1-(-\sqrt{3)}(\sqrt{3})} \end{aligned} $$

$$ \begin{array}{rlrl} \text { and } & & \frac{y+2}{x-3} & =\frac{-\sqrt{3}-\sqrt{3}}{1+(-\sqrt{3})(\sqrt{3})} \\ \Rightarrow & & \frac{y+2}{x-3} & =0 \\ \text { and } & & =\frac{-2 \sqrt{3}}{1-3}=\sqrt{3} \\ y+2 & =\sqrt{3} x-3 \sqrt{3} \end{array} $$

Neglecting, $y+2=0$, as it does not intersect $Y$-axis.



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