# Find the equation of a straight line on which the perpendicular from the origin makes an angle of 30° with x–axis and which forms a triangle of the area with the axes.

Assuming: AB be the given line, and OL = p be the perpendicular drawn from the origin on the line.

Given: α = 60°

Explanation:

So, the equation of the line AB is

x cos θ + y sin θ = p

x cos 30 + y sin 30 = p

√3x + y = 2p …… (1)

Now, in triangles OLA and OLB

Cos 30° = , cos6° =

,

OA = and OB = 2p

It is given that the area of triangle OAB is 503

p2 = 75

p = √75

Substituting the value of p in (1)

√3 x + y = √75

Hence, the equation of the line AB is √3 x + y = √75

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