Find the equation of the line on which the length of the perpendicular segment from the origin to the line is 4 and the inclination of the perpendicular segment with the positive direction of x-axis is $30^{\circ} .$

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Solution

$The normal form of the equation of the line is x c o s ω + y s i n ω = p . Here p = 4 a n d ω = 30^{\circ} . Therefore, the equation of the line is$

$x c o s 30^{\circ} + y s i n 30^{\circ} = 4$

$x \frac{\sqrt{3}}{2} + y \frac{1}{2} = 4 o r \sqrt{3} x + y = 8$

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Find the equation of the line on which the length of the perpendicular segment from the origin to the line is 4 and the inclination of the perpendicular segment with the positive direction of x-axis is 30∘.

The normal form of the equation of the line is x cos ω+y sin ω=p. Here p=4 and ω=30∘. Therefore, the equation of the line is x cos 30∘+y sin 30∘=4 x√32+y12=4 or √3 x+y=8

Find the equation of the line on which the length of the perpendicular segment from the origin to the line is 4 and the inclination of the perpendicular segment with the positive direction of x-axis is $30^{\circ} .$

$The normal form of the equation of the line is x c o s ω + y s i n ω = p . Here p = 4 a n d ω = 30^{\circ} . Therefore, the equation of the line is$

$x c o s 30^{\circ} + y s i n 30^{\circ} = 4$

$x \frac{\sqrt{3}}{2} + y \frac{1}{2} = 4 o r \sqrt{3} x + y = 8$