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:
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.
Watch out
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
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.
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:
L1:2x+3y−5=0
L2:x−y+2=0
L3:3x+2y−3=0
Method 1 (substitution):
Solve L1 and L2:
From L2: x=y−2
Substitute into L1: 2(y−2)+3y−5=0⇒2y−4+3y−5=0⇒5y=9⇒y=59
Then x=59−2=−51
Intersection point: (−51,59)
Now check L3: 3(−51)+2(59)−3=−53+518−3=515−3=3−3=0
Concurrent lines = three or more lines meeting at one point.
Condition: The determinant of their coefficients (including constants) must be zero.
Check: Also ensure no two lines are parallel (otherwise the determinant trick can mislead you).
That is the whole idea — from the visual of ropes meeting at a flagpole to the algebraic test that tells you instantly whether they do.
The Concurrent Lines Condition is a standard determinant-based test from the NCERT Class 11 Mathematics chapter on Straight Lines, and it directly answers searches like "condition for three lines to be concurrent" or "straight lines important questions class 11". This determinant-equals-zero check is also a frequently tested shortcut in JEE Main and various state CET coordinate geometry questions.
Concept: Concurrent Lines Condition — three lines are concurrent if the point of intersection of any two lies on the third.
Step 1: Solve the first and third equations for intersection.
From 2x+y−3=0 and 3x−y−2=0, add them:
5x−5=0⇒x=1.
Substitute x=1 into 2(1)+y−3=0⇒y=1.
So intersection point is (1,1).
Step 2: For concurrency, (1,1) must satisfy the second line 5x+ky−3=0.
Substitute: 5(1)+k(1)−3=0⇒5+k−3=0⇒k+2=0.
Step 3: Hence k=−2.
✓Final answer
The value is −2.
For three lines to be concurrent, they must all pass through a single common point. The value of k is found by first solving any two lines for their intersection, then substituting that point into the third line. The required value is k=−2.
Why the Concurrent Lines Condition Works
Three lines are concurrent when they all meet at exactly one point. This means the intersection point of any two lines must also lie on the third line. So the strategy is simple: find where two of the lines cross, then force the third line to pass through that same point.
A common mistake is to try using the determinant condition for concurrency straight away — that works too, but it's more mechanical and less intuitive. The substitution method is cleaner and shows you exactly what's happening geometrically.
Step-by-step solution
1. Pick two lines to find their intersection.
The simplest pair to solve is the first and third lines:
2x+y−33x−y−2=0(1)=0(3)
Add them directly — the y terms cancel:
(2x+3x)+(y−y)+(−3−2)=0
5x−5=0
x=1
2. Find the corresponding y value.
Substitute x=1 into equation (1):
2(1)+y−3=0
2+y−3=0
y−1=0
y=1
So the intersection point of lines (1) and (3) is (1,1).
Tip
Always check your intersection with the other line you used. Put (1,1) into equation (3): 3(1)−1−2=0 — it works. This confirms you haven't made an arithmetic slip.
3. Force the second line to pass through this point.
The second line is:
5x+ky−3=0
For concurrency, (1,1) must satisfy it:
5(1)+k(1)−3=0
5+k−3=0
k+2=0
k=−2
Watch out
A common error is to forget the sign when moving terms. Here 5−3=2, so k+2=0 gives k=−2, not k=2. Always isolate k carefully.
4. Verify (optional but good practice).
With k=−2, the second line becomes 5x−2y−3=0. Check (1,1): 5−2−3=0. All three lines now pass through (1,1), so they are concurrent.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
KCET 2025Set A-11 markMCQ
Q.The equation of the line through the point (0,1,2) and perpendicular to the line 2x−1=3y+1=−2z−1 is
(A) 3x=4y−1=−3z−2
(B) −3x=4y−1=3z−2
(C) 3x=4y−1=3z−2
(D) 3x=−4y−1=3z−2
›Reveal solutionSolution
Two lines are perpendicular iff the dot product of their direction ratios is zero — test each option's ⟨a,b,c⟩ against ⟨2,3,−2⟩.
Step 1 — Extract the direction ratios of the given line
2x−1=3y+1=−2z−1⟹d1=⟨2,3,−2⟩
Step 2 — The perpendicularity condition
All four options already pass through (0,1,2) (each has the form ax−0=by−1=cz−2), so the point condition does not discriminate. What must discriminate is the angle.
For the required line with direction ratios d2=⟨a,b,c⟩, perpendicularity means
d1⋅d2=0⟹2a+3b−2c=0
Step 3 — Test each option
Option
⟨a,b,c⟩
2a+3b−2c
Perpendicular?
(A)
⟨3,4,−3⟩
6+12+6=24
✗
(B)
⟨−3,4,3⟩
−6+12−6=0
✓
(C)
⟨3,4,3⟩
6+12−6=12
✗
(D)
⟨3,−4,3⟩
6−12−6=−12
✗
Step 4 — Confirm the point
Option (B) is −3x−0=4y−1=3z−2; substituting (0,1,2) gives 0=0=0✓, so the line indeed passes through (0,1,2).
Both conditions — passing through (0,1,2)and perpendicular to the given line — are met only by (B).
✓Final answer
The correct option is (B) — −3x=4y−1=3z−2.
ANSWER: B
KCET 2024Set A-11 markMCQ
Q.The angle between the line x+y=3 and the line joining the points (1,1) and (−3,4) is
(A) tan−1(7)
(B) tan−1(−71)
(C) tan−1(71)
(D) tan−1(72)
›Reveal solutionSolution
Get both slopes, then apply the angle-between-two-lines formula tanθ=1+m1m2m1−m2.
Step 1 — Slope of the first line.x+y=3⇒y=−x+3, so
m1=−1.
Step 2 — Slope of the line joining (1,1) and (−3,4).
m2=x2−x1y2−y1=−3−14−1=−43=−43.
Step 3 — Angle between them. The acute angle θ between two lines of slopes m1,m2 (with 1+m1m2=0) satisfies
tanθ=1+m1m2m1−m2.
This comes from θ=θ1−θ2 where mi=tanθi, together with the subtraction formula for tan.
Substituting:
m1−m2=−1−(−43)=−1+43=−41,
1+m1m2=1+(−1)(−43)=1+43=47.
Hence
tanθ=47−41=41⋅74=71.
Step 4 — Conclude.
θ=tan−1(71).
(Option (B), tan−1(−1/7), would be the obtuse supplement — but the angle between two lines is taken as the acute one, so the modulus is used.)
✓Final answer
The correct option is (C) — tan−1(71).
ANSWER: C
KCET 2023Set A-21 markMCQ
Q.If x[32]+y[1−1]=[155] then the value of x and y are
(A) x=4,y=−3
(B) x=−4,y=−3
(C) x=−4,y=3
(D) x=4,y=3
›Reveal solutionSolution
This is a system of two linear equations in two unknowns, written in vector form. Solving it gives x=4 and y=3, which corresponds to option (D).
The problem gives you a vector equation:
x[32]+y[1−1]=[155].
When two vectors are equal, each corresponding component must be equal. So this single vector equation is really two separate equations — one from the top row and one from the bottom row. That’s the core idea: break the vector equality into a system of scalar equations, then solve.
Write the component equations.
From the first (top) component: 3x+y=15.
From the second (bottom) component: 2x−y=5.
Solve the system.
Add the two equations to eliminate y:
(3x+y)+(2x−y)=15+5
5x=20
x=4.
Substitute x=4 into the first equation:
3(4)+y=15
12+y=15
y=3.
Check with the second equation: 2(4)−3=8−3=5, which matches.
Watch out
A common mistake is to forget that vector equality means component-wise equality. Another is to mis-handle the signs when adding or subtracting the equations — always write them out clearly.
Tip
You can also solve by treating this as a matrix equation [321−1][xy]=[155] and using the inverse matrix, but elimination is faster here.
✓Final answer
The correct option is (D): x=4, y=3.
COMEDK 2021Set 2021-B1 markMCQ
Q.If the lines ax+2y+1=0, bx+3y+1=0 and cx+4y+1=0 are concurrent, then which of the following relationships are correct?
(A) 2b=a+c
(B) b=a+c
(C) b2=ac
(D) a+b+c=0
›Reveal solutionSolution
Concurrency requires 2b=a+c.
The lines ax+2y+1=0, bx+3y+1=0, cx+4y+1=0 are concurrent iff
abc234111=0.
Expanding along the first row:
a(3−4)−2(b−c)+1(4b−3c)=−a−2b+2c+4b−3c=−a+2b−c=0.
Hence 2b=a+c.
✓Final answer
The correct option is (A) — 2b=a+c
KCET 2020Set A-11 markMCQ
Q.If the vectors 2i^−3j^+4k^, 2i^+j^−k^ and λi^−j^+2k^ are coplanar, then the value of λ is
(A) 6
(B) −5
(C) −6
(D) 5
›Reveal solutionSolution
Coplanar ⇔ scalar triple product [abc]=0; set the determinant to zero and solve for λ.
Step 1 — The concept
The scalar triple product [abc]=a⋅(b×c) equals (up to sign) the volume of the parallelepiped spanned by the three vectors. If the three vectors lie in a single plane, that parallelepiped is flat — zero volume. Hence
a,b,c are coplanar⟺a1b1c1a2b2c2a3b3c3=0
Check: with λ=6, the rows are (2,−3,4), (2,1,−1), (6,−1,2). Determinant =2(2−1)+3(4+6)+4(−2−6)=2+30−32=0✓ — the volume really is zero, so the three vectors are coplanar.