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: Find the slope.
The line makes an angle $90^\circ+\alpha$ with the X-axis, so its slope is $m=\tan(90^\circ+\alpha)$.
Step 2: Simplify the slope.
Since $\tan(90^\circ+\alpha)=-\cot\alpha$, we have $m=-\dfrac{\cos\alpha}{\sin\alpha}$.
Step 3: Use point-slope form.
The line passes through $(p\cos\alpha,\,p\sin\alpha)$, so $y-p\sin\alpha=-\dfrac{\cos\alpha}{\sin\alpha}\,(x-p\cos\alpha)$.
Step 4: Clear the denominator.
Multiply both sides by $\sin\alpha$: $y\sin\alpha-p\sin^2\alpha=-x\cos\alpha+p\cos^2\alpha$.
Step 5: Collect the x and y terms.
Bring them together: $x\cos\alpha+y\sin\alpha=p\cos^2\alpha+p\sin^2\alpha$.
Step 6: Use the Pythagorean identity.
Since $\cos^2\alpha+\sin^2\alpha=1$, the right side is just $p$. So $x\cos\alpha+y\sin\alpha=p$, which is option (C).
\[ \boxed{\,x\cos\alpha+y\sin\alpha=p\,} \]