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

To find the equation of the line in its new position, we can break down the process into clear, logical steps.

Step 1: Determine the initial inclination and rotation of the line
Initially, the line passes through the point $A(9,0)$ and makes an angle of $30°$ with the positive direction of the x-axis.
When the line is rotated about $A$ through an angle of $15°$ in the clockwise direction, the new angle of inclination, let's call it $\theta$, is reduced. Since clockwise rotation decreases the angle of inclination with the positive x-axis:
$\theta =30°-15°=15°$

Step 2: Find the slope of the line in its new position
The slope $m$ of the line in its new position is given by:
$m=\tan (15°)$

We can compute $\tan (15°)$ using the tangent subtraction formula:
$\tan (15°)=\tan (45°-30��)=\frac{\tan (45°)-\tan (30°)}{1+\tan (45��)\tan (30°)}$

Substituting the known values $\tan (45°)=1$ and $\tan (30°)=\frac{1}{\sqrt{3}}$:
$m=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$

To simplify this value, we rationalize the denominator:
$m=\frac{(\sqrt{3}-1{)}^2}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{3+1-2\sqrt{3}}{3-1}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}$

Step 3: Write the equation of the line
Using the point-slope form, the equation of the line passing through $A(9,0)$ with slope $m=2-\sqrt{3}$ is:
$y-0=(2-\sqrt{3})(x-9)$
$y=(2-\sqrt{3})(x-9)$

Let us rearrange this to match the given options. Dividing both sides by $2-\sqrt{3}$:
$\frac{y}{2-\sqrt{3}}=x-9$

Multiplying the numerator and denominator of the left-hand side fraction by -1, we get:
$\frac{y}{\sqrt{3}-2}=-(x-9)$
$\frac{y}{\sqrt{3}-2}=-x+9$

Rearranging the terms, we obtain:
$\frac{y}{\sqrt{3}-2}+x=9$

Therefore, the correct option matches the equation:
$\frac{y}{\sqrt{3}-2}+x=9$