Hence, equation of required line using normal form is
xcos15∘+ysin15∘=4 or xcos75∘+ysin75∘=4
⇒(3+1)x+(3−1)y=82 or (3−1)x+(3+1)y=82
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∘ with the line x+y=0. Then an equation of the line L is:
Note: In actual JEE Main paper, two options were correct for this question. Hence, we have changed one option.
Held on 12 Apr 2019 · Verified 6 Jul 2026.
(3+1)x+(3−1)y=82
x+3y=8
3x+y=8
(3−1)x+3y=82
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