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 50√3
⇒
⇒ 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