The set of lines ax+by+c = 0, where 3a + 2b + 4c = 0 is concurrent at the point .........

Q: Wait, why do we divide by 4? What if the coefficient of c was zero?

Great question! We divide by 4 because we need the coefficient of $c$ in our constraint to match the coefficient of $c$ in the general line equation, which is $1$. If the coefficient of $c$ were zero, the lines would be parallel, and they would not be concurrent at a finite point (or they would be concurrent at infinity). In this specific problem, the $4$ is the key to unlocking the coordinates.

Q: Why can't we just pick random values for a and b and solve for c?

You absolutely could! If you pick two different sets of values for $a$ and $b$ that satisfy the constraint, you will get two different lines. Since all lines in the family pass through the same point, the intersection of these two specific lines will be the point of concurrency. However, the algebraic method of comparing coefficients is much faster and more elegant for JEE problems.

Q: Is there a geometric interpretation for this?

Yes! The constraint $3a + 2b + 4c = 0$ essentially restricts the 'degrees of freedom' of the line. Instead of having three independent parameters, you only have two. Geometrically, this means the lines are not just any lines; they are a specific 'pencil' of lines that are forced to pass through a common intersection point. You are essentially finding the intersection of two lines from this family.

Question

The set of lines $ax+by+c = 0$, where $3a + 2b + 4c = 0$ is concurrent at the point .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing in a vast, empty coordinate plane. You have the power to draw an infinite number of lines, each defined by the equation $ax + by + c = 0$.

Without any restrictions, these lines could be anywhere, pointing in any direction, creating a chaotic web of geometry. But then, we introduce a constraint: $3a + 2b + 4c = 0$.

Suddenly, the chaos vanishes. The lines are no longer free; they are bound by a hidden law. They are now a 'family' of lines, and they all share a secret, magical property: they are concurrent.

This means every single line in this family, no matter how you choose $a$, $b$, and $c$, passes through one single, fixed point. Our mission is to find this point.

### The Logic of the Lock

To find this point, let us call it $(x_1, y_1)$, we must realize that if it lies on every line, it must satisfy the equation $ax_1 + by_1 + c = 0$ for any valid $a, b, c$.

We are given the constraint $3a + 2b + 4c = 0$. This is our key.

If we can make this constraint look exactly like the line equation, we can simply read off the coordinates.

### The Normalization: The Aha! Moment

Here is where most students stumble. They look at $ax + by + c = 0$ and $3a + 2b + 4c = 0$ and try to compare them immediately.

But wait! The coefficient of $c$ in the first equation is $1$, while in the second, it is $4$. We have a mismatch.

To fix this, we must normalize the constraint. We divide the entire equation $3a + 2b + 4c = 0$ by $4$.

This gives us:

$$\frac{3a}{4} + \frac{2b}{4} + \frac{4c}{4} = 0$$

Which simplifies beautifully to:

$$\frac{3}{4}a + \frac{1}{2}b + c = 0$$

Now, compare this with $x(a) + y(b) + c = 0$. The structure is identical! By matching the coefficients of $a$ and $b$, we see that $x = \frac{3}{4}$ and $y = \frac{1}{2}$.

### The Final Triumph

We have done it! The point of concurrency is **$(\frac{3}{4}, \frac{1}{2})$**.

Every line that satisfies the constraint $3a + 2b + 4c = 0$ will pass through this exact point. It is a beautiful example of how a simple algebraic manipulation can reveal a deep geometric truth.

Remember, in JEE Advanced, it is rarely about brute force; it is about finding the right perspective, normalizing your equations, and letting the symmetry reveal the answer. Keep practicing, keep visualizing, and most importantly, keep falling in love with the elegance of mathematics.

Animated Solution for Mathematics - Straight Lines: The set of lines $a x + b y + c = 0$ , where $3 a + 2 b + 4 c = 0$ is concurrent at the point .........

Visualized Solution

The Problem Setup

The Family of Lines

The Concept of Concurrency

The Concurrency Condition

Analyzing the Given Constraint

The Mismatch in Constant Term

Normalizing the Constraint

The Simplified Constraint

Extracting the Coordinates

The Final Conclusion

The Sigma Insight: Family of Lines

Analyzing the Setup

The Logic of the Lock

The Normalization

The Final Triumph

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Animated Solution for Mathematics - Straight Lines: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The set of lines ax+by+c=0, where 3a+2b+4c=0 is concurrent at the point .........