Question

Equation of a straight line which makes an angle \(150^\circ\) with the positive direction of X-axis and an intercept \(4\) units on the Y-axis is

• A. \(x+y\sqrt{2}=4\sqrt{2}\)
• B. \(y+x\sqrt{3}=4\)
• C. \(x+y\sqrt{3}=4\sqrt{3}\)
• D. \(x-y\sqrt{3}=4\sqrt{3}\)

Hint:

If a line makes an angle \(\theta\) with the positive X-axis, then \[ m=\tan\theta. \] The slope-intercept form is \[ y=mx+c, \] where \(c\) is the Y-intercept.

Answer 1

The Correct Option is C

Step 1: Find the slope of the line. The angle made with the positive \(X\)-axis is \[ 150^\circ. \] Hence, \[ m=\tan150^\circ =-\frac{1}{\sqrt3}. \]

Step 2: Use the given Y-intercept. The line cuts the Y-axis at \(4\), so it passes through \((0,4)\). Using point-slope form, \[ y-4=-\frac{1}{\sqrt3}x. \]

Step 3: Simplify the equation. Multiplying by \(\sqrt3\), \[ \sqrt3\,y-4\sqrt3=-x, \] or \[ x+\sqrt3\,y=4\sqrt3. \]

Step 4: Final conclusion. \[ \boxed{x+\sqrt3\,y=4\sqrt3} \] Hence, the correct option is \(\boxed{(C)}\).