Q. 104.3( 3 Votes )

# 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.

Answer :

**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

⇒

⇒ p^{2} = 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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