Straight Line And Pair Of Straight Lines Ques 17
- 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:
A straight line passing through $P$ and making an angle of $\alpha=60^{\circ}$, is given by $y - y_1 = m(x - x_1)$ where $m = \tan(\alpha)$
$$ \frac{y-y _1}{x-x _1}=\tan (\theta \pm \alpha) $$
$$ \Rightarrow \quad \sqrt{3} x+y=1 $$
$\Rightarrow y=-\sqrt{3} x+1$, then $\tan \theta=-\sqrt{3}$
$$ \begin{aligned} \frac{y+2}{x-3} & =\frac{\tan \theta \pm \tan \alpha}{1 \mp \tan \theta \tan \alpha} \\ \frac{y+2}{x-3} & =\frac{-\sqrt{3}+\sqrt{3}}{1-(-\sqrt{3)}(\sqrt{3})} \end{aligned} $$
$$ \begin{array}{rlrl} \text { and } & & \frac{y+2}{x-3} & =\frac{-\sqrt{3}-\sqrt{3}}{1+(-\sqrt{3})(\sqrt{3})} \\ \Rightarrow & & \frac{y+2}{x-3} & =0 \\ and & & =\frac{-2 \sqrt{3}}{1-3}=-\sqrt{3} \\ y+2 & =\sqrt{3} x-3 \sqrt{3} \end{array} $$
Neglecting, $y+2=0$, as it does not intersect the $y$-axis.
BETA
