Q.If three lines whose equations are $y = m_1 x + c_1$, $y = m_2 x + c_2$ and $y = m_3 x + c_3$ are concurrent, then show that $m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0$.

What Does "Concurrent Lines" Mean?

Imagine three friends standing in a field. Each friend holds a long, straight rope stretched tight. If all three ropes pass through exactly the same point — say, a flagpole in the centre — then the ropes are concurrent. That common point is called the point of concurrency.

In geometry, when three or more lines all pass through a single point, we say they are concurrent lines. That point is their concurrency point.

The Intuition Behind the Condition

Suppose you have three lines:

• $L_1: a_1x + b_1y + c_1 = 0$
• $L_2: a_2x + b_2y + c_2 = 0$
• $L_3: a_3x + b_3y + c_3 = 0$

If they are concurrent, there exists some point $(x_0, y_0)$ that satisfies all three equations at once. That means $(x_0, y_0)$ is a common solution.

Now, think about it this way:

The first two lines $L_1$ and $L_2$ intersect at some point $P$ (unless they are parallel). For the three lines to be concurrent, $L_3$ must also pass through that same point $P$. So the condition boils down to: the point of intersection of any two lines must lie on the third line.

That is the simplest way to check concurrency: solve two equations, get the intersection, and plug it into the third equation. If it satisfies, the lines are concurrent.

The Precise Algebraic Condition

There is a cleaner, more powerful condition using determinants — it avoids solving for the intersection explicitly.

This determinant being zero is the necessary and sufficient condition for concurrency of three lines.

Why Does the Determinant Work?

Here is the reasoning in plain steps:

1. For concurrency, there must exist $(x_0, y_0)$ such that:

2. Think of these as three equations in three unknowns: $x_0$, $y_0$, and the constant $1$. Yes, the constant 1 is treated as a variable here.

3. For a non-trivial solution to exist (i.e., a solution where the "variables" are not all zero), the determinant of the coefficient matrix must be zero. That is a standard result from linear algebra: a homogeneous system has a non-zero solution only when the determinant is zero.

4. The determinant being zero is exactly the condition that the three equations are linearly dependent — meaning one equation can be written as a combination of the other two. That is another way to say: the third line passes through the intersection of the first two.

Example

Check if these lines are concurrent:

$L_1: 2x + 3y - 5 = 0$

$L_2: x - y + 2 = 0$

$L_3: 3x + 2y - 3 = 0$

Method 1 (substitution):

Solve $L_1$ and $L_2$: …