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JEE Main 201912 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
12 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Straight Lines

JEE Main 2019 · 12 April · Evening Shift · Mathematics · Straight Lines

A straight line L at a distance of 4 units from the origin makes...

A straight line LL at a distance of 44 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°60 ° with the line x+y=0.x + y = 0 . Then an equation of the line LL is:
Note: In actual JEE Main paper, two options were correct for this question. Hence, we have changed one option.A straight line L at a distance of 4 units from the origin makes positive intercepts on the coordinate axes and the

Options

  1. A(3+1)x+(3−1)y=82\left(\sqrt{3} + 1\right)x + \left(\sqrt{3} - 1\right)y = 8\sqrt{2}
  2. Bx+3y=8x + \sqrt{3}y = 8
  3. C3x+y=8\sqrt{3}x + y = 8
  4. D(3−1)x+3y=82\left(\sqrt{3} - 1\right)x + \sqrt{3}y = 8\sqrt{2}

Correct answer

A. (3+1)x+(3−1)y=82\left(\sqrt{3} + 1\right)x + \left(\sqrt{3} - 1\right)y = 8\sqrt{2}

Step-by-step solution

A straight line L at a distance of 4 units from the origin makes positive intercepts on the coordinate axes and the
Hence, equation of required line using normal form is
xcos15°+ysin15°=4xcos{15}^{°} + ysin{15}^{°} = 4 or xcos75°+ysin75°=4xcos{75}^{°} + ysin{75}^{°} = 4
⇒(3+1)x+(3−1)y=82⇒ ( \sqrt{3} + 1 ) x + ( \sqrt{3} - 1 ) y = 8\sqrt{2} or (3−1)x+(3+1)y=82( \sqrt{3} - 1 ) x + ( \sqrt{3} + 1 ) y = 8\sqrt{2}
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