Q

# Help me understand! - Co-ordinate geometry - JEE Main-7

The 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 $60^{\circ}$ with the line

$x+y=0$ . Then an equation of the line L is :

• Option 1)

$x+\sqrt3y=8$

• Option 2)

$(\sqrt3+1)x+(\sqrt3-1)y=8\sqrt2$

• Option 3)

$\sqrt3x+y=8$

• Option 4)

$(\sqrt3-1)x+(\sqrt3+1)y=8\sqrt2$

Views

Hence, the equation of line is

$xcos\theta+ysin\theta=p$

$xcos75^{\circ}+ysin75^{\circ}=4$

$x(\frac{\sqrt3-1}{2\sqrt2})+y(\frac{\sqrt3+1}{2\sqrt2})=4$

$(\sqrt3+1)x+(\sqrt3-1)y=8\sqrt2$

Option 1)

$x+\sqrt3y=8$

Option 2)

$(\sqrt3+1)x+(\sqrt3-1)y=8\sqrt2$

Option 3)

$\sqrt3x+y=8$

Option 4)

$(\sqrt3-1)x+(\sqrt3+1)y=8\sqrt2$

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