Question

The equation of a line passing through $(p \cos \alpha, p \sin \alpha)$ and making an angle $(90 + \alpha)$ with positive direction of X-axis is

• A. $x \cos \alpha - y \sin \alpha = 2p$
• B. $x \sin \alpha + y \cos \alpha = p$
• C. $x \cos \alpha + y \sin \alpha = p$
• D. $x \cos \alpha + y \sin \alpha = 3p$

Hint:

The normal form of a line is $x \cos \theta + y \sin \theta = p$, where $p$ is the perpendicular distance from the origin. The point $(p \cos \alpha, p \sin \alpha)$ is the point where the perpendicular from origin meets the line.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need the equation of a line passing through a specific point with a given inclination.

Step 2: Key Formula or Approach:
The slope $m = \tan \theta$. Given $\theta = 90^\circ + \alpha$, $m = \tan(90^\circ + \alpha) = -\cot \alpha = -\frac{\cos \alpha}{\sin \alpha}$. Use the point-slope form $y - y_1 = m(x - x_1)$.

Step 3: Detailed Explanation:
Equation: $y - p \sin \alpha = -\frac{\cos \alpha}{\sin \alpha} (x - p \cos \alpha)$.
Multiplying by $\sin \alpha$: $y \sin \alpha - p \sin^2 \alpha = -x \cos \alpha + p \cos^2 \alpha$.
Rearranging: $x \cos \alpha + y \sin \alpha = p \cos^2 \alpha + p \sin^2 \alpha$.
Since $\cos^2 \alpha + \sin^2 \alpha = 1$, we get $x \cos \alpha + y \sin \alpha = p$.

Step 4: Final Answer:
The equation of the line is $x \cos \alpha + y \sin \alpha = p$, which is option (C).