Find the equation of the line which is at a distance 3 from the origin and the perpendicular from the origin to the line makes an angle of $30^{\circ}$ with the positive direction of the x axis

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Solution

Here p = 3, $α = 30^{\circ}$
$∴$ Equation of the line in the normal form is
$x cos 30^{\circ} + y sin 30^{\circ} = 3$ or $x \frac{\sqrt{3}}{2} + \frac{y}{2} = 3$ or $\sqrt{3} x + y = 6$

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Find the equation of the line which is at a distance 3 from the origin and the perpendicular from the origin to the line makes an angle of 30∘ with the positive direction of the x axis

Here p = 3, α=30∘∴ Equation of the line in the normal form isxcos30∘+ysin30∘=3 or x√32+y2=3 or √3x+y=6

Find the equation of the line which is at a distance 3 from the origin and the perpendicular from the origin to the line makes an angle of $30^{\circ}$ with the positive direction of the x axis

Here p = 3, $α = 30^{\circ}$
$∴$ Equation of the line in the normal form is
$x cos 30^{\circ} + y sin 30^{\circ} = 3$ or $x \frac{\sqrt{3}}{2} + \frac{y}{2} = 3$ or $\sqrt{3} x + y = 6$