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: Get the slope.
The line makes angle $90^{\circ}+\alpha$ with the X-axis, so its slope is $\tan(90^{\circ}+\alpha) = -\cot\alpha = -\tfrac{\cos\alpha}{\sin\alpha}$.

Step 2: Use point-slope form.
Through $(p\cos\alpha, p\sin\alpha)$: \[ y - p\sin\alpha = -\tfrac{\cos\alpha}{\sin\alpha}(x - p\cos\alpha). \]

Step 3: Clear the fraction.
Multiply through by $\sin\alpha$: $y\sin\alpha - p\sin^2\alpha = -x\cos\alpha + p\cos^2\alpha$. Bring the $x$ and $y$ terms together: $x\cos\alpha + y\sin\alpha = p(\cos^2\alpha + \sin^2\alpha)$.

Step 4: Simplify.
Since $\cos^2\alpha + \sin^2\alpha = 1$, the line is \[ \boxed{x\cos\alpha + y\sin\alpha = p} \]