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 $60^{\circ}$ with the line $x + y = 0.$ Then an equation of the line $L$ is :

(\sqrt{3} + 1) x + (\sqrt{3} - 1) y = 8 \sqrt{2}

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(\sqrt{3} - 1) x + (\sqrt{3} + 1) y = 8 \sqrt{2}

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\sqrt{3} x + y = 8

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x + \sqrt{3} y = 8

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Solution

The correct option is B $(\sqrt{3} - 1) x + (\sqrt{3} + 1) y = 8 \sqrt{2}$

$O A = p = 4$
$x + y = 0$ makes an angle of $135^{\circ}$ with the positive $x$ -axis.

Equation of the line $L$ in perpendicular form is
$x cos θ + y sin θ = p \Rightarrow x cos 75^{\circ} + y sin 75^{\circ} = p \Rightarrow x (\frac{\sqrt{3} - 1}{2 \sqrt{2}}) + y (\frac{\sqrt{3} + 1}{2 \sqrt{2}}) = 4 \Rightarrow x (\sqrt{3} - 1) + y (\sqrt{3} + 1) = 8 \sqrt{2}$

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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 :

The correct option is B (√3−1)x+(√3+1)y=8√2 OA=p=4 x+y=0 makes an angle of 135∘ with the positive x-axis. Equation of the line L in perpendicular form is xcosθ+ysinθ=p⇒xcos75∘+ysin75∘=p⇒x(√3−12√2)+y(√3+12√2)=4⇒x(√3−1)+y(√3+1)=8√2

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 $60^{\circ}$ with the line $x + y = 0.$ Then an equation of the line $L$ is :