A straight line L through the point (3, -2) is inclined at an angle 60° to the line √3x + y = 1. If

Q: Wait, why are there two possible slopes? Shouldn't a line be unique?

That is a brilliant observation! In geometry, a line is indeed unique once defined by a point and a slope. However, the condition 'inclined at $60^\circ$' is symmetric. Imagine the reference line as a mirror; you can tilt your new line $60^\circ$ clockwise or $60^\circ$ counter-clockwise. Both satisfy the condition, which is why the modulus in our formula gives us two distinct paths.

Q: Why did we reject the $m=0$ case? It's a valid slope, isn't it?

You are absolutely right that $m=0$ is a valid mathematical slope. However, in JEE Advanced, we must always respect the 'physical' constraints of the problem. The problem explicitly states that the line must intersect the x-axis. A line with $m=0$ passing through $(3, -2)$ is $y = -2$, which is a horizontal line parallel to the x-axis. It will never touch the x-axis, so we must discard it.

Q: Is there a faster way to do this without the slope formula?

You could use vector normals or rotation matrices, but for this specific problem, the slope formula is the most efficient 'surgical' tool. It cuts straight to the answer without the overhead of vector algebra. Stick to the slope formula for lines—it is your best friend in coordinate geometry.

Question

A straight line $L$ through the point $(3, -2)$ is inclined at an angle $60^\circ$ to the line $\sqrt{3}x + y = 1$. If $L$ also intersects the x-axis, then the equation of $L$ is

Options:

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

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

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

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

Sigma Percentile · Accepted Answer

### The Geometry of Symmetry: Unlocking the Mystery Line

Welcome, future engineer. Today, we are not just solving a line equation; we are exploring the symmetry of the coordinate plane.

When you look at a problem like this, it is easy to get lost in the algebra. But I want you to pause and visualize. We have a point $P(3, -2)$ and a reference line $L_1$ defined by $\sqrt{3}x + y = 1$.

We are looking for a line $L$ that passes through $P$ and cuts $L_1$ at exactly $60^\circ$. Imagine standing at point $P$. You have a compass; you draw a line at $60^\circ$ to the right, and another at $60^\circ$ to the left. Both are valid, and the algebra will give us two answers, but the geometry will force us to choose the correct one.

### Phase 1: The Slope of the Reference

Before we can dance with the angles, we need to know the 'rhythm' of our reference line. The equation $\sqrt{3}x + y = 1$ is currently in a standard form, but it hides its true nature.

Let us rearrange it into the slope-intercept form, $y = mx + c$. By subtracting $\sqrt{3}x$ from both sides, we get $y = -\sqrt{3}x + 1$.

Now, the slope $m_1$ reveals itself clearly as $m_1 = -\sqrt{3}$. This is the anchor for our entire calculation.

### Phase 2: The Modulus Trap

Now, we invoke the most powerful tool in our coordinate geometry arsenal: the angle formula between two lines. We know that:

$$	an 	heta = \left| \frac{m - m_1}{1 + m m_1} ight|$$

Here, $	heta = 60^\circ$, so $	an 60^\circ = \sqrt{3}$. Substituting our known slope $m_1 = -\sqrt{3}$, the equation becomes:

$$\sqrt{3} = \left| \frac{m - (-\sqrt{3})}{1 + m(-\sqrt{3})} ight| \Rightarrow \sqrt{3} = \left| \frac{m + \sqrt{3}}{1 - \sqrt{3}m} ight|$$

Do not fear the modulus! It is simply telling us that there are two directions in which we can incline our line. We must solve for both.

### Phase 3: The Branching Paths

Let us split this into two cases.

**Case 1:** 
$$\sqrt{3} = \frac{m + \sqrt{3}}{1 - \sqrt{3}m}$$

Cross-multiplying, we get $\sqrt{3}(1 - \sqrt{3}m) = m + \sqrt{3}$. Expanding this, we find $\sqrt{3} - 3m = m + \sqrt{3}$. The $\sqrt{3}$ terms cancel out beautifully, leaving us with $4m = 0$, or $m = 0$.

**Case 2:** 
$$-\sqrt{3} = \frac{m + \sqrt{3}}{1 - \sqrt{3}m}$$

Cross-multiplying gives $-\sqrt{3} + 3m = m + \sqrt{3}$. Rearranging, we get $2m = 2\sqrt{3}$, which simplifies to $m = \sqrt{3}$.

### Phase 4: The Geometric Filter

We have two candidates for our slope: $m = 0$ and $m = \sqrt{3}$. But wait! The problem imposes a strict constraint: the line must intersect the x-axis.

If we choose $m = 0$, our line passing through $(3, -2)$ becomes $y - (-2) = 0(x - 3)$, which simplifies to $y = -2$. This is a horizontal line parallel to the x-axis. It will never intersect it, so we must reject this.

Thus, our only valid slope is $m = \sqrt{3}$.

### The Final Victory

With our slope $m = \sqrt{3}$ and our point $(3, -2)$, we use the point-slope form: $y - y_1 = m(x - x_1)$.

Substituting our values, we get $y - (-2) = \sqrt{3}(x - 3)$. Expanding this, we have $y + 2 = \sqrt{3}x - 3\sqrt{3}$.

Bringing everything to one side, we arrive at the final equation:

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

You have navigated the modulus, respected the geometric constraints, and arrived at the truth. This is the essence of JEE Advanced—precision, logic, and a deep respect for the geometry behind the algebra.

Animated Solution for Mathematics - Straight Lines: A straight line $L$ through the point $(3, - 2)$ is inclined at an angle $6 0^{\circ}$ to the line $3 x + y = 1$ . If $L$ also intersects the x-axis, then the equation of $L$ is

Select Answer:

Visualized Solution

Visualizing the Problem

Slope of Given Line $L_{1}$

The Angle Formula

Setting up the Equation

Solving for $m$ : Case 1

Solving for $m$ : Case 2

Applying the Constraint

Final Equation of $L$

Conclusion and Key Takeaway

The Sigma Insight: Angle Between Two Lines

The Geometry of Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

The Final Victory

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