Question

The equation of a line, having inclination 120° with positive direction of x-axis, which is at a distance of 3 units from the origin is​

Answer 1

Answer:

$\sqrt{3}.x+y=6$

Step-by-step explanation:

Let us assume that the straight line is AB which makes 120° with the positive X-axis direction. {See the diagram}

So, ∠ABX=120°

Then, ∠ABO=180°-120°=60°

Now, draw a perpendicular line on AB from origin O, which is OC (say) .

From ΔOCB, ∠OCB=90° and ∠CBO=60°,  

then ∠COB=180°-60°-90°=30°...... (1)

Given that the straight line AB is at 3 units distance from the origin.

Then, OC=3 units .....(2)

Now, we have a formula that the equation of a straight line is given by

xcosα+ysinα=p ...... (3)

where α is the angle with the positive X-axis, made by the perpendicular line drawn from origin to that straight line and p is the perpendicular distance of that line from origin.

Now, it is clear from diagram that, here α=∠COB=30° {From equation (1)} and p=OC=3 units {From equation (2)}.

Now, from equation (3), the equation of the straight line will be

xcos30°+ysin30°=3

⇒$\frac{\sqrt{3} x}{2}+\frac{y}{2}=3$

⇒$\sqrt{3}.x+y=6$ (Answer)