MHT CET · Maths · Straight Lines
The equation of line, where length of the perpendicular segment from origin to the line is 4 and the inclination of this perpendicular segment with the positive direction of \(\mathrm{X}\)-axis is \(30^{\circ}\), is
- A \(x+\sqrt{3} y=8\)
- B \(x-\sqrt{3} y=8\)
- C \(\sqrt{3} x-y=8\)
- D \(\sqrt{3} x+y=8\)
Answer & Solution
Correct Answer
(D) \(\sqrt{3} x+y=8\)
Step-by-step Solution
Detailed explanation
We have \(\ell(\mathrm{OP})=4\) and \(\mathrm{m} \angle \mathrm{POX}=30^{\circ}\)
\(\therefore \mathrm{P} \equiv\left(4 \cos 30^{\circ}, 4 \sin 30^{\circ}\right) \equiv(2 \sqrt{3}, 2)\)
From figure, we conclude that angle made be line with +ve \(\mathrm{X}\) axis is \(120^{\circ}\).
\(\therefore\) Slope of line \(=\tan \left(120^{\circ}\right)=-\sqrt{3}\)
Hence required equation of line \(\mathrm{L}\) is
\((y-2)=(-\sqrt{3})(x-2 \sqrt{3}) \Rightarrow \sqrt{3} x+y=8\)
\(\therefore \mathrm{P} \equiv\left(4 \cos 30^{\circ}, 4 \sin 30^{\circ}\right) \equiv(2 \sqrt{3}, 2)\)
From figure, we conclude that angle made be line with +ve \(\mathrm{X}\) axis is \(120^{\circ}\).
\(\therefore\) Slope of line \(=\tan \left(120^{\circ}\right)=-\sqrt{3}\)
Hence required equation of line \(\mathrm{L}\) is
\((y-2)=(-\sqrt{3})(x-2 \sqrt{3}) \Rightarrow \sqrt{3} x+y=8\)
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