If a,b,c are in A.P., then the straight lines ax+by+c=0 will always pass through __.

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a,b,c

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Concept understanding — Concurrent Lines Condition

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.

Note

Two lines are always concurrent (unless they are parallel) — they meet at exactly one point. The interesting case is three or more lines. Do they all happen to pass through the same spot?

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.

a_1x + b_1y + c_1 = 0

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

Watch out

This condition assumes that no two of the lines are parallel. If $L_1$ and $L_2$ are parallel, they never meet, so the three lines cannot be concurrent (unless all three are the same line, which is a degenerate case). The determinant condition will still give zero in that parallel case, but the lines are not concurrent — they are parallel. So always check that the lines actually intersect pairwise first.

Why Does the Determinant Work?

Here is the reasoning in plain steps:

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

a_1x_0 + b_1y_0 + c_1 = 0

a_2x_0 + b_2y_0 + c_2 = 0

a_3x_0 + b_3y_0 + c_3 = 0

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.
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.
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.

Tip

For quick checks in exams, use the determinant. But if the numbers are simple, solving two equations and substituting into the third is often faster and less error-prone.

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 $L_2$ : …

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The condition that $a,b,c$ are in A.P. allows us to rewrite the equation of the line $ax+by+c=0$ in a form that reveals it always passes through a fixed point. This point is $\boxed{(1, -2)}$ .

When we say that a family of straight lines "always passes through" a certain point, it means that the coordinates of this point satisfy the equation of every line in that family, regardless of the specific values of the parameters defining the family (in this case, $a, b, c$ ), as long as those parameters meet the given condition.

The key idea here is to use the condition that $a,b,c$ are in A.P. to eliminate one of the parameters from the line equation, or to express the equation in a form that highlights a fixed point. If we can rewrite the equation $ax+by+c=0$ as $P(x,y) \cdot \lambda_1 + Q(x,y) \cdot \lambda_2 = 0$ , where $\lambda_1$ and $\lambda_2$ are parameters (or related to $a,b,c$ ) and $P(x,y)$ and $Q(x,y)$ are expressions involving $x$ and $y$ , then for this equation to hold true for any valid $\lambda_1, \lambda_2$ , the point $(x,y)$ must satisfy both $P(x,y)=0$ and $Q(x,y)=0$ . This intersection point is the fixed point.

Let's apply this to the problem.

Express the A.P. condition:

Given that $a,b,c$ are in A.P., the defining property is that the middle term is the average of the other two, or equivalently, the difference between consecutive terms is constant.

This means:

$b - a = c - b$

Rearranging this, we get:

$2b = a+c$
Substitute the A.P. condition into the line equation:

We have the equation of the straight line:

$ax+by+c=0$

From the A.P. condition, we can express $b$ in terms of $a$ and $c$ : $b = \frac{a+c}{2}$ .

Substitute this into the line equation:

$ax + \left(\frac{a+c}{2}\right)y + c = 0$
Rearrange and group terms:

To clear the fraction, multiply the entire equation by 2:

$2ax + (a+c)y + 2c = 0$

Now, distribute $y$ and group terms containing $a$ and terms containing $c$ :

$2ax + ay + cy + 2c = 0$

$a(2x+y) + c(y+2) = 0$ …

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Q.If a,b,ca,b,ca,b,c are in A.P., then the straight lines ax+by+c=0ax+by+c=0ax+by+c=0 will always pass through ____.

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Q.If $a,b,c$ are in A.P., then the straight lines $ax+by+c=0$ will always pass through ____.

Unlock everything free for 14 days