A) \[y+2=0,\ \ \sqrt{3}x-y-2-3\sqrt{3}=0\]
B) \[x-2=0,\ \ \sqrt{3}x-y+2+3\sqrt{3}=0\]
C) \[\sqrt{3}x-y-2-3\sqrt{3}=0\]
D) None of these
Correct Answer: A
Solution :
The equation of any straight line passing through (3, ?2) is \[y+2=m(x-3)\] ?..(i) The slope of the given line is \[-\sqrt{3}\]. So, \[\tan {{60}^{o}}=\pm \frac{m-(-\sqrt{3})}{1+m\text{ }(-\sqrt{3})}\] On solving, we get \[m=0\] or \[\sqrt{3}\] Putting the values of m in (i), the required equation of lines are \[y+2=0\]and \[\sqrt{3}x-y=2+3\sqrt{3}\].
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