Question

The equation of line, where length of the perpendicular segment from origin to the line is 4 and the inclination of this perpendicular segment with the positive direction of X-axis is $30^\circ$, is

• A. $x + \sqrt{3}y = 8$
• B. $x - \sqrt{3}y = 8$
• C. $\sqrt{3}x - y = 8$
• D. $\sqrt{3}x + y = 8$

Hint:

Whenever a problem specifies "perpendicular distance from the origin" and the "inclination of that perpendicular," bypass all slope calculations and immediately drop the numbers into the standard formula $x \cos \alpha + y \sin \alpha = p$.

Answer 1

The Correct Option is D

Step 1: Identify the type of data given.
We know the length of the perpendicular from the origin to the line is $p = 4$ and its inclination with the positive $x$-axis is $\alpha = 30^\circ$. This is exactly the information the normal form uses.
Step 2: Recall the normal form.
A line at perpendicular distance $p$ from the origin, with the perpendicular making angle $\alpha$ with the $x$-axis, is $x\cos\alpha + y\sin\alpha = p$.
Step 3: Plug in the angle.
Substituting $\alpha = 30^\circ$ and $p = 4$ gives $x\cos 30^\circ + y\sin 30^\circ = 4$.
Step 4: Insert the trig values.
Using $\cos 30^\circ = \dfrac{\sqrt{3}}{2}$ and $\sin 30^\circ = \dfrac{1}{2}$, we get $\dfrac{\sqrt{3}}{2}x + \dfrac{1}{2}y = 4$.
Step 5: Combine over a common denominator.
The left side becomes $\dfrac{\sqrt{3}x + y}{2} = 4$.
Step 6: Clear the fraction.
Multiplying both sides by 2 yields $\sqrt{3}x + y = 8$, which is option 4 and agrees with the key.
\[ \boxed{\sqrt{3}x + y = 8} \]