Question

If \(\alpha\) is the angle made by the perpendicular drawn from origin to the line \(12x - 5y + 13 = 0\) with the positive X-axis in anti-clockwise direction, then \(\alpha =\)

• A. \(\operatorname{Tan}^{-1} \frac{5}{12} \)
• B. \(2\pi - \operatorname{Tan}^{-1} \frac{5}{12} \)
• C. \(\pi - \operatorname{Tan}^{-1} \frac{5}{12} \)
• D. \(\pi + \operatorname{Tan}^{-1} \frac{5}{12} \)

Hint:

To find the angle \(\alpha\) of the normal from the origin to a line \(Ax + By + C = 0\): 1. Rewrite the equation as \(Ax + By = -C\). 2. Ensure the constant term on the right-hand side is positive. If \(-C\) is negative, multiply the entire equation by \(-1\). 3. Divide the entire equation by \(\sqrt{A^2 + B^2}\) (using the \(A\) and \(B\) from the equation after adjusting for positive constant). 4. The coefficients of \(x\) and \(y\) will be \(\cos \alpha\) and \(\sin \alpha\) respectively. 5. Determine the quadrant of \(\alpha\) based on the signs of \(\cos \alpha\) and \(\sin \alpha\), then find \(\alpha\) using the inverse tangent function and quadrant rules.

Answer 1

The Correct Option is C

Step 1: Convert the given line equation to the normal form.
The equation of the line is \(12x - 5y + 13 = 0\).
The general form of a linear equation is \(Ax + By + C = 0\). Here, \(A = 12\), \(B = -5\), \(C = 13\).
The normal form of a line is \(x \cos \alpha + y \sin \alpha = p\), where \(p\) is the perpendicular distance from the origin to the line, and \(\alpha\) is the angle that this perpendicular makes with the positive x-axis. To convert \(Ax + By + C = 0\) to normal form, we divide the entire equation by \(\pm \sqrt{A^2 + B^2}\). The sign is chosen such that the constant term \(p\) (which is \(-C / \pm \sqrt{A^2 + B^2}\)) is positive. First, calculate \(\sqrt{A^2 + B^2}\): \[ \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \] The given equation is \(12x - 5y + 13 = 0\).
To make the constant term positive in the normal form, we need to rewrite the equation as \(-12x + 5y = 13\).
Now, divide by 13: \[ \frac{-12}{13}x + \frac{5}{13}y = \frac{13}{13} \] \[ \frac{-12}{13}x + \frac{5}{13}y = 1 \] Step 2: Identify the values of \(\cos \alpha\) and \(\sin \alpha\).
Comparing this equation with the normal form \(x \cos \alpha + y \sin \alpha = p\), we get: \[ \cos \alpha = -\frac{12}{13} \] \[ \sin \alpha = \frac{5}{13} \] And \(p = 1\). 
Step 3: Determine the angle \(\alpha\).
We observe that \(\cos \alpha\) is negative and \(\sin \alpha\) is positive.
This indicates that the angle \(\alpha\) lies in the second quadrant.
Let \(\theta\) be the acute angle such that \(\tan \theta = \left| \frac{\sin \alpha}{\cos \alpha} \right|\). \[ \tan \theta = \left| \frac{5/13}{-12/13} \right| = \left| -\frac{5}{12} \right| = \frac{5}{12} \] So, \(\theta = \operatorname{Tan}^{-1} \left( \frac{5}{12} \right)\). Since \(\alpha\) is in the second quadrant, and \(\theta\) is the reference angle in the first quadrant, \(\alpha\) can be expressed as: \[ \alpha = \pi - \theta \] \[ \alpha = \pi - \operatorname{Tan}^{-1} \left( \frac{5}{12} \right) \] The final answer is $\boxed{\pi - \operatorname{Tan}^{-1} \frac{5}{12}}$.