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Misc-I · Q89

Q.The equation of a line, having inclination 120° with positive direction of X−axis, which is at a distance of 3 units from the origin is (A) √3x ± y+6=0 (B) √3x + y±6=0 (C) x+y=6 (D) x+y=−6

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A line of inclination 120°120° has its normal from the origin at angle α=120°−90°=30°\alpha=120°-90°=30° (or the supplementary case). Using the normal form xcos⁡α+ysin⁡α=px\cos\alpha+y\sin\alpha=p with p=3,α=30°p=3,\alpha=30°: 32x+12y=3⇒3x+y=6\dfrac{\sqrt3}{2}x+\dfrac12y=3 \Rightarrow \sqrt3x+y=6. Allowing for the line's …

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