JEE MAIN - Mathematics (2019 - 12th April Evening Slot - No. 6)
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 :
x + $$\sqrt 3 $$y = 8
$$\sqrt 3 $$x + y = 8
( $$\sqrt 3 $$ + 1)x + ( $$\sqrt 3 $$ – 1)y = 8 $$\sqrt 2 $$
( $$\sqrt 3 $$ - 1)x + ( $$\sqrt 3 $$ + 1)y = 8 $$\sqrt 2 $$
Explanation
_12th_April_Evening_Slot_en_6_1.png)
The equation of line is
x cos $$\theta $$ + y sin $$\theta $$ = p
$$ \Rightarrow $$ x cos (75o) + y sin (75o) = 4
$$ \Rightarrow $$ $$x\left( {{{\sqrt 3 - 1} \over {2\sqrt 2 }}} \right) + y\left( {{{\sqrt 3 + 1} \over {2\sqrt 2 }}} \right) = 4$$
$$ \Rightarrow $$ $$x\left( {\sqrt 3 - 1} \right) + y\left( {\sqrt 3 + 1} \right) = 8\sqrt 2 $$
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