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QJEEM19Maf14a: JEE Main 2019 Mathematics - Class 11 JEEM19Maf14a Solution: A straight line L at a distance of 4 units from the origin…

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 60o with the line x + y = 0. Then an equation of the line L is :

Multiple Choice Options

  • A x + 3\sqrt 3 y = 8
  • B 3\sqrt 3 x + y = 8
  • C ( 3\sqrt 3 + 1)x + ( 3\sqrt 3 – 1)y = 8 2\sqrt 2
  • D ( 3\sqrt 3 - 1)x + ( 3\sqrt 3 + 1)y = 8 2\sqrt 2

Step-by-Step Verified Solution

JEE Main 2019 (Online) 12th April Evening Slot Mathematics - Straight Lines and Pair of Straight Lines Question 96 English Explanation

The equation of line is
x cos θ\theta + y sin θ\theta = p

⇒\Rightarrow x cos (75o) + y sin (75o) = 4

⇒\Rightarrow x(3−122)+y(3+122)=4x\left( {{{\sqrt 3 - 1} \over {2\sqrt 2 }}} \right) + y\left( {{{\sqrt 3 + 1} \over {2\sqrt 2 }}} \right) = 4

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

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