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:
- $\frac{x}{\sqrt{3} + 2} + y = 9 \\$
- $\frac{y}{\sqrt{3} - 2} + x = 9$
- $\frac{x}{\sqrt{3} + 2} + y = 9 \\$
- $\frac{x}{\sqrt{3} - 2} + y = 9$
The Correct Option is B
Approach Solution - 1
The problem asks for the equation of a line that initially passes through the point \( A(9, 0) \) with an angle of inclination of \( 30^\circ \), after it is rotated about point \( A \) by \( 15^\circ \) in the clockwise direction.
Concept Used:
The solution requires the following concepts from coordinate geometry:
- Slope of a Line: The slope (\(m\)) of a line that makes an angle \( \theta \) with the positive x-axis is given by \( m = \tan(\theta) \).
- Rotation of a Line: When a line with an initial angle of inclination \( \theta_{initial} \) is rotated clockwise by an angle \( \alpha \), the new angle of inclination is \( \theta_{new} = \theta_{initial} - \alpha \).
- Point-Slope Form: The equation of a line passing through a point \( (x_1, y_1) \) with a slope \( m \) is given by the formula \( y - y_1 = m(x - x_1) \).
We will also use the trigonometric identity for the tangent of a difference of angles: \( \tan(A - B) = \frac{\tan(A) - \tan(B)}{1 + \tan(A)\tan(B)} \).
Step-by-Step Solution:
Step 1: Determine the new angle of inclination of the line after rotation.
The initial angle of the line with the positive x-axis is \( \theta_{initial} = 30^\circ \).
The line is rotated by \( 15^\circ \) in the clockwise direction. A clockwise rotation decreases the angle of inclination. Therefore, the new angle is:
\[ \theta_{new} = \theta_{initial} - 15^\circ = 30^\circ - 15^\circ = 15^\circ \]Step 2: Calculate the slope (\(m\)) of the line in its new position.
The slope of the new line is given by \( m = \tan(\theta_{new}) = \tan(15^\circ) \).
We can find the value of \( \tan(15^\circ) \) using the angle subtraction formula:
\[ m = \tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan(45^\circ) - \tan(30^\circ)}{1 + \tan(45^\circ)\tan(30^\circ)} \]Substituting the standard values \(\tan(45^\circ) = 1\) and \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\):
\[ m = \frac{1 - \frac{1}{\sqrt{3}}}{1 + (1)\left(\frac{1}{\sqrt{3}}\right)} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \]To simplify, we rationalize the denominator:
\[ m = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3 + 1 - 2\sqrt{3}}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \]Step 3: Use the point-slope form to find the equation of the new line.
The line passes through the point of rotation \( A(x_1, y_1) = (9, 0) \) and has a slope of \( m = 2 - \sqrt{3} \).
\[ y - y_1 = m(x - x_1) \] \[ y - 0 = (2 - \sqrt{3})(x - 9) \] \[ y = (2 - \sqrt{3})(x - 9) \]Final Computation & Result:
Step 4: Check which of the given options matches the derived equation.
The derived equation is \( y = (2 - \sqrt{3})(x - 9) \).
Let's examine option (1):
\[ \frac{y}{\sqrt{3}-2} + x = 9 \]Rearranging this equation to solve for \(y\):
\[ \frac{y}{\sqrt{3}-2} = 9 - x \] \[ y = (\sqrt{3}-2)(9 - x) \]Factoring out a -1 from both terms:
\[ y = - (2 - \sqrt{3}) \times -(x - 9) \] \[ y = (2 - \sqrt{3})(x - 9) \]This matches our derived equation. Therefore, option (2) is the correct answer.
The equation of the line in the new position is \(\frac{y}{\sqrt{3}-2} + x = 9\).
Approach Solution -2
Expanding and rearranging leads to the equation \( \frac{y}{\sqrt{3} - 2} + x = 9 \), which matches Option (2).
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