Question

A line passing through the point \( A(9, 0) \) makes an angle of \( 30^\circ \) with the positive direction of the x-axis. If this line is rotated about \( A \) through an angle of \( 15^\circ \) in the clockwise direction, then its equation in the new position is:

• A. $\frac{x}{\sqrt{3} + 2} + y = 9 \\$
• B. $\frac{y}{\sqrt{3} - 2} + x = 9$
• C. $\frac{x}{\sqrt{3} + 2} + y = 9 \\$
• D. $\frac{x}{\sqrt{3} - 2} + y = 9$

Answer 1

The Correct Option is B

The problem requires determining the equation of a line. Initially, the line passes through point \( A(9, 0) \) and has an angle of inclination of \( 30^\circ \). This line is subsequently rotated \( 15^\circ \) clockwise around point \( A \).

Core Concepts:

The solution relies on the following principles from coordinate geometry:

1. Slope Calculation: A line's slope (\(m\)) is determined by the tangent of its angle of inclination (\( \theta \)) with the positive x-axis: \( m = \tan(\theta) \).
2. Angle Rotation: A clockwise rotation by an angle \( \alpha \) from an initial angle \( \theta_{initial} \) results in a new angle \( \theta_{new} = \theta_{initial} - \alpha \).
3. Point-Slope Form: The equation of a line passing through \( (x_1, y_1) \) with slope \( m \) is \( y - y_1 = m(x - x_1) \).

Additionally, the tangent of a difference formula, \( \tan(A - B) = \frac{\tan(A) - \tan(B)}{1 + \tan(A)\tan(B)} \), will be utilized.

Detailed Solution:

Step 1: Calculate the new angle of inclination post-rotation.

The line's initial angle of inclination is \( \theta_{initial} = 30^\circ \). A clockwise rotation of \( 15^\circ \) modifies this angle.

\[\theta_{new} = 30^\circ - 15^\circ = 15^\circ\]

Step 2: Compute the slope (\(m\)) of the rotated line.

The slope is \( m = \tan(\theta_{new}) = \tan(15^\circ) \).

Using the angle subtraction formula:

\[m = \tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan(45^\circ) - \tan(30^\circ)}{1 + \tan(45^\circ)\tan(30^\circ)}\]

With \(\tan(45^\circ) = 1\) and \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\):

\[m = \frac{1 - \frac{1}{\sqrt{3}}}{1 + (1)\left(\frac{1}{\sqrt{3}}\right)} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}\]

Rationalizing the denominator yields:

\[m = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3 + 1 - 2\sqrt{3}}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}\]

Step 3: Apply the point-slope form to derive the new line's equation.

The line passes through \( A(x_1, y_1) = (9, 0) \) with slope \( m = 2 - \sqrt{3} \).

\[y - 0 = (2 - \sqrt{3})(x - 9)\]\[y = (2 - \sqrt{3})(x - 9)\]

Final Calculation and Outcome:

Step 4: Compare the derived equation with the provided options.

The derived equation is \( y = (2 - \sqrt{3})(x - 9) \).

Consider option (1):

\[\frac{y}{\sqrt{3}-2} + x = 9\]

Rearranging to isolate \(y\):

\[\frac{y}{\sqrt{3}-2} = 9 - x\]\[y = (\sqrt{3}-2)(9 - x)\]

Factoring out -1:

\[y = - (2 - \sqrt{3}) \times -(x - 9)\]\[y = (2 - \sqrt{3})(x - 9)\]

This confirms option (2) as the correct match. The equation of the line in its new orientation is \(\frac{y}{\sqrt{3}-2} + x = 9\).