A straight line L through the point $(3,-2)$ is inclined at an angle of $60^{\circ}$ to the line $\sqrt{3}…

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

Solution

The equation of a straight line passing through $(3,-2)$ is $y+2=m(x-3)$
The slope of the line $\sqrt{3} x+y=1$ is $-\sqrt{3}$ So, $\tan 60^{\circ}= \pm \frac{m-(-\sqrt{3})}{1+m(-\sqrt{3})} \Rightarrow \sqrt{3}= \pm \frac{m+\sqrt{3}}{1-\sqrt{3} m}$ On solving, we get $\mathrm{m}=0 \text { or } \sqrt{3}$
Putting the values of m in (i), the required equation of lines are $y+2=0$ and $y-\sqrt{3} x+2+3 \sqrt{3}=0$

Asked in: MHT CET 2024 (11 May Shift 2)

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