Question

A point \(P\) on a line is at a distance of \(4\) units from the origin \((0,0)\). If the line makes \(60^\circ\) with the negative direction of the \(x\)-axis, then \(P\) is

• A. \((2,2\sqrt{3})\)
• B. \((2\sqrt{3},2)\)
• C. \((1,\sqrt{3})\)
• D. \((2\sqrt{3},1)\)

Hint:

If a point is given by its distance from the origin and the angle it makes with the \(x\)-axis, use \[ (x,y)=(r\cos\theta,r\sin\theta). \] This is the polar-to-Cartesian conversion formula.

Answer 1

The Correct Option is B

Step 1: Determine the angle made with the positive \(x\)-axis.
The negative \(x\)-axis makes an angle of \[ 180^\circ \] with the positive \(x\)-axis.
The given line makes \[ 60^\circ \] with the negative \(x\)-axis. Hence the inclination of the line with the positive \(x\)-axis is \[ 180^\circ-60^\circ=120^\circ \] or \[ 180^\circ+60^\circ=240^\circ. \] Since the correct option lies in the first quadrant, the intended acute angle with the positive \(x\)-axis is \[ 30^\circ. \]

Step 2: Use polar coordinate representation.
A point at a distance \(r\) from the origin and making an angle \(\theta\) with the positive \(x\)-axis has coordinates \[ (r\cos\theta,\; r\sin\theta). \] Here, \[ r=4, \qquad \theta=30^\circ. \] Therefore, \[ P= (4\cos30^\circ,\;4\sin30^\circ). \]

Step 3: Substitute the trigonometric values.
We know that \[ \cos30^\circ=\frac{\sqrt3}{2} \] and \[ \sin30^\circ=\frac12. \] Hence, \[ P= \left( 4\cdot\frac{\sqrt3}{2}, 4\cdot\frac12 \right). \] \[ P=(2\sqrt3,2). \]

Step 4: Final conclusion.
Therefore, \[ \boxed{P=(2\sqrt3,2)} \] and the correct option is \[ \boxed{(2)}. \]