A line passing through the point A(9, 0) makes an angle of 30° with the positive direction of x-axis

Q: Wait, why does clockwise rotation mean subtracting the angle?

In coordinate geometry, the angle of inclination $\theta$ is measured counter-clockwise from the positive x-axis. If you rotate a line clockwise, you are effectively moving 'down' towards the x-axis, which reduces the angle of inclination. Hence, $\theta_{new} = \theta_{initial} - \text{rotation angle}$.

Q: Do I really need to memorize the value of $\tan(15^\circ)$?

Not at all! While it is a handy value to know, the JEE rewards the ability to derive it. Using $\tan(45^\circ - 30^\circ)$ is the standard, elegant way to ensure you don't make a sign error during the exam.

Q: My final equation looks different from the options. Did I mess up?

Don't panic! Algebraic expressions can take many forms. If you have $y = (2 - \sqrt{3})(x - 9)$, try dividing both sides by $(2 - \sqrt{3})$ and then multiplying the numerator and denominator by the conjugate if necessary. The options often use specific signs to test your comfort with rationalization.

Q: Why is the point $(9, 0)$ so important?

That point is your 'pivot'. In geometry, a line is defined by a point and a direction. By fixing the pivot at $(9, 0)$, you ensure that no matter how the slope changes, the line must always pass through that specific coordinate.

Q: Is there a shortcut using rotation matrices?

You could use rotation matrices, but for a simple line rotation about a point, the point-slope form is significantly faster and less prone to calculation errors. Always choose the path of least resistance in the exam!

Question

A line passing through the point $A(9, 0)$ makes an angle of $30^\circ$ with the positive direction of $x$-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

Options:

(A) $\frac{y}{\sqrt{3}-2} + x = 9$

(B) $\frac{x}{\sqrt{3}-2} + y = 9$

(C) $\frac{x}{\sqrt{3}+2} + y = 9$

(D) $\frac{y}{\sqrt{3}+2} + x = 9$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast Cartesian plane. You are anchored at the point $A(9, 0)$, a fixed sentinel on the x-axis.

A line passes through you, stretching out into the distance at an angle of $30^{\circ}$ relative to the positive x-axis. This is your starting state.

We are tasked with a rotation—a $15^{\circ}$ clockwise shift. Let us embark on this transformation together.

### Defining the New Direction

When we talk about the 'angle of inclination' $	heta$, we are talking about the line's orientation relative to the positive x-axis. A clockwise rotation is a movement that brings the line closer to the x-axis.

If we start at $	heta_1 = 30^{\circ}$ and rotate clockwise by $15^{\circ}$, our new angle of inclination becomes $	heta_2 = 30^{\circ} - 15^{\circ} = 15^{\circ}$.

The slope $m$ is defined as $	an(	heta)$. Thus, we must find $m = 	an(15^{\circ})$.

### The Elegance of Trigonometry

Calculating $	an(15^{\circ})$ is best achieved using the compound angle formula. We know that $15^{\circ} = 45^{\circ} - 30^{\circ}$.

Using the identity $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$, we substitute our values:

$$
	an(15^{\circ}) = \frac{	an(45^{\circ}) - 	an(30^{\circ})}{1 + 	an(45^{\circ})	an(30^{\circ})} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}}
$$

By multiplying the numerator and denominator by $\sqrt{3}$, we get $\frac{\sqrt{3} - 1}{\sqrt{3} + 1}$. To make this usable, we rationalize the denominator by multiplying by the conjugate $(\sqrt{3} - 1)$:

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

This value, $2 - \sqrt{3}$, is the 'soul' of our new line. It dictates exactly how steep our path is.

### Constructing the Equation

Now that we have our slope $m = 2 - \sqrt{3}$ and our pivot point $(x_1, y_1) = (9, 0)$, we invoke the point-slope form: $y - y_1 = m(x - x_1)$.

Substituting our values, we obtain:

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

To match standard JEE formats, we rearrange the terms. Dividing by $(2 - \sqrt{3})$, we get:

$$
\frac{y}{2 - \sqrt{3}} = x - 9
$$

We note that $2 - \sqrt{3} = -(\sqrt{3} - 2)$. Substituting this into our denominator yields:

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

Moving the negative sign to the numerator and transposing the terms, we arrive at the **final, elegant form**:

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

### Conclusion: The Beauty of the Process

We started with a simple line, applied a geometric transformation, utilized trigonometric identities, and performed algebraic rationalization to reach a precise destination.

Mathematics is not just about finding the answer; it is about the logical flow from one state to the next. You have successfully navigated the rotation, and in doing so, you have mastered another piece of the coordinate geometry puzzle.

Select Answer:

Visualized Solution

Initial Position of the Line

Clockwise Rotation

New Angle of Inclination

Evaluating $tan (1 5^{\circ})$

Rationalizing the Slope

Point-Slope Form

Substituting the Values

Rearranging the Equation

Matching with Options

Final Equation

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

Defining the New Direction

The Elegance of Trigonometry

Constructing the Equation

Conclusion

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