Question

Let the line \( l_1 \) be a line passing through the point \( (0,-6) \) and making an angle of \(150^\circ\) with the positive \(x\)-axis. The equation of a line \( l_2 \) parallel to \( l_1 \) and crossing the \(y\)-axis at a distance \(2\) units below the origin is:

• A. \(x+\sqrt3y+2\sqrt3=0\)
• B. \(x-\sqrt3y-2\sqrt3=0\)
• C. \(\sqrt3x+y+6=0\)
• D. \(x-\sqrt3y+6\sqrt3=0\)

Hint:

Parallel lines always have the same slope. First determine the slope from the angle, then use the point-slope form of a line.

Answer 1

The Correct Option is A

Understanding the Concept: The slope of a line making angle \(\theta\) with the positive \(x\)-axis is: \[ m=\tan\theta \] Parallel lines always have equal slopes.
Step 1: Find the slope of line \(l_1\). Given: \[ \theta=150^\circ \] Therefore, \[ m=\tan150^\circ \] Using the identity: \[ \tan(180^\circ-\theta)=-\tan\theta \] we get: \[ \tan150^\circ=-\tan30^\circ \] \[ =-\frac1{\sqrt3} \] Thus the slope is: \[ m=-\frac1{\sqrt3} \]
Step 2: Determine the point through which \(l_2\) passes. The line crosses the \(y\)-axis \(2\) units below the origin. Therefore, the point is: \[ (0,-2) \]
Step 3: Use point-slope form. Equation of line: \[ y-y_1=m(x-x_1) \] Substituting: \[ y+2=-\frac1{\sqrt3}(x-0) \] \[ \sqrt3y+2\sqrt3=-x \] Bringing all terms to one side: \[ x+\sqrt3y+2\sqrt3=0 \] Hence, \[ \boxed{x+\sqrt3y+2\sqrt3=0} \]