JEE Mains · Maths · STD 11 - 9. straight line
A straight line \(L\) at a distance of \(4\) units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of \(60^o\) with the line \(x + y = 0\). Then an equation of the line \(L\) is
- A \(\left( {\sqrt 3 - 1} \right)x + \left( {\sqrt 3 + 1} \right)y = 4\sqrt 2 \)
- B \(\sqrt 3 x + y = 8\)
- C \(x + \sqrt 3 y = 8\)
- D \(\left( {\sqrt 3 + 1} \right)x + \left( {\sqrt 3 - 1} \right)y = 8\sqrt 2 \)
Answer & Solution
Correct Answer
(D) \(\left( {\sqrt 3 + 1} \right)x + \left( {\sqrt 3 - 1} \right)y = 8\sqrt 2 \)
Step-by-step Solution
Detailed explanation
\(OP = 4\) given \(OP\) makes \({60^o}\) with \(x+y=0\) Let slope of \(OP=m\) \( \Rightarrow \tan \,{60^o} = \left| {\frac{{m + 1}}{{1 - m}}} \right|\) \( \Rightarrow \frac{{m + 1}}{{1 - m}} = \sqrt 3 \) or \( - \sqrt 3 \) \(m + 1 = \sqrt {3m} - 3\) or…
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