AP EAMCET · Maths · Straight Lines
A straight line \(\mathrm{L}\) at a distance of 4 units from the origin makes positive intercepts on the coordinate axes and the perpendicular drawn from the origin to this line makes an angle of \(60^{\circ}\) with the line \(x+y=0\). Then the equation of the line \(\mathrm{L}\) is
- A \((\sqrt{3}+1) x+(\sqrt{3}-1) y=8 \sqrt{2}\)
- B \((\sqrt{3}-1) x+(\sqrt{3}+1) y=8 \sqrt{2}\)
- C \(\sqrt{3} x+y=8\)
- D \(x+\sqrt{3} y=8\)
Answer & Solution
Correct Answer
(B) \((\sqrt{3}-1) x+(\sqrt{3}+1) y=8 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
Given: \(O P=4 \mathrm{~cm}\) Slope of \(y+x=0\) is -1 . Slope of \(O P\) is \(\alpha\). Then, \(60+\alpha=135^{\circ} \Rightarrow \alpha=75^{\circ}\) Now, \(\cos 75^{\circ}=\cos \left(30^{\circ}+45^{\circ}\right)=\frac{\sqrt{3}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}}\)…
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