Q.The lines ax+2y+1=0, bx+3y+1=0 and cx+4y+1=0 are concurrent if a,b,c are in G.P.
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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.
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:
- L1:a1x+b1y+c1=0
- L2:a2x+b2y+c2=0
- L3:a3x+b3y+c3=0
If they are concurrent, there exists some point (x0,y0) that satisfies all three equations at once. That means (x0,y0) is a common solution.
Now, think about it this way:
The first two lines L1 and L2 intersect at some point P (unless they are parallel). For the three lines to be concurrent, L3 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.
Three lines a1x+b1y+c1=0, a2x+b2y+c2=0, a3x+b3y+c3=0 are concurrent if and only if
a1a2a3b1b2b3c1c2c3=0
This determinant being zero is the necessary and sufficient condition for concurrency of three lines.
This condition assumes that no two of the lines are parallel. If L1 and L2 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 (x0,y0) such that:
a1x0+b1y0+c1=0
a2x0+b2y0+c2=0
a3x0+b3y0+c3=0
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Think of these as three equations in three unknowns: x0, y0, and the constant 1. Yes, the constant 1 is treated as a variable here.
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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.
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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.
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:
L1:2x+3y−5=0
L2:x−y+2=0
L3:3x+2y−3=0
Method 1 (substitution):
Solve L1 and L2: …
Concept: Three lines are concurrent if they meet at a single point, which happens when the determinant of their coefficients vanishes.
The three lines are concurrent when:
abc234111=0
Expanding along the first column:
a(3−4)−b(2−4)+c(2−3)=0
−a+2b−c=0
This gives us 2b=a+c.
Since a,b,c are in G.P., we have b2=ac (the defining property of a geometric progression).
From 2b=a+c, squaring both sides: 4b2=a2+2ac+c2. …
The concurrency determinant reduces to 2b=a+c, i.e. a,b,c are in A.P. — NOT G.P. — so the given statement is incorrect.
Solution
The lines ax+2y+1=0, bx+3y+1=0, cx+4y+1=0 are concurrent if and only if
abc234111=0.
Expand the determinant:
a(3−4)−2(b−c)+(4b−3c)=−a−2b+2c+4b−3c=−a+2b−c.
Set it equal to zero:
−a+2b−c=0 ⇒ 2b=a+c. …
- KEAM 2025Set eng-2025-04264 marksMCQQ.Let ax+by+c=0 the equation of a straight line such that 3a+2b+4c=0. Which one of the following points, lies on the line? (A) (43,21) (B) (21,43) (C) (41,23) (D) (23,21) (E) (2,4)
›Reveal solutionSolution
A line ax+by+c=0 whose coefficients satisfy 3a+2b+4c=0 passes through the fixed point obtained by matching (x,y,1)∝(3,2,4), giving (43,21). …
- KEAM 2025Set eng-2025-04274 marksMCQQ.Which one of the following lines, passes through the point of intersection of x+y=5 and 2x+y=7? (A) 4x+3y=−1 (B) 3x+2y=7 (C) 4x−3y=−1 (D) 4x+3y−2=0 (E) 4x+3y+3=0
›Reveal solutionSolution
The intersection point is (2,3); substituting it, only line 4x−3y=−1 holds.
Solve the system: subtracting x+y=5 from 2x+y=7 gives x=2, then y=3. Intersection (2,3).
Test options:
- 4(2)+3(3)=17=−1 …
- KEAM 2022Set eng-2022-P2-B14 marksMCQQ.If the lines 2x−3y+5=0, 9x−5y+14=0 and 3x−7y+λ=0 are concurrent, then the value of λ is equal to (A) 7 (B) 8 (C) 10 (D) 9 (E) 6
›Reveal solutionSolution
λ=10.
Concept and Intuition
Three lines are concurrent if they share a common point; find that point from two lines, then force the third through it.
Step-by-Step Solution
- From 2x−3y=−5 and 9x−5y=−14: multiply to eliminate y — 10x−15y=−25, 27x−15y=−42; subtract: 17x=−17⇒x=−1.
- 2(−1)−3y=−5⇒y=1. Point (−1,1). …
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