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.
The three lines intersect at a common point when p=5. This is found by first solving the intersection of two lines, then substituting that point into the third line to satisfy the concurrent lines condition.
The Core Idea: What Does "Intersect at One Point" Mean?
When three lines are given, they can either form a triangle (three distinct intersection points), be parallel in various ways, or all pass through a single common point. The phrase "intersect at one point" means the three lines are concurrent — they all meet at the same point.
The most reliable method: find where any two lines meet, then force the third line to pass through that same point. This works because if three lines are concurrent, the intersection of any two must lie on the third.
Watch out
A common mistake is to try solving all three equations simultaneously at once. That can work, but it's messier. The cleaner approach is to solve two equations first, then substitute into the third.
Step-by-Step Solution
1. Pick two lines and find their intersection.
The simplest pair to solve is the first and third lines:
3x+y−2=0 → y=2−3x
2x−y−3=0 → y=2x−3
Set them equal:
2−3x=2x−3
2. Solve for x.
Bring terms together:
2+3=2x+3x
5=5x
x=1
3. Find the corresponding y.
Substitute x=1 into either equation. Using y=2−3(1):
y=2−3=−1
So the intersection point of the first and third lines is (1,−1).
Tip
Always check your point in the other equation as a quick sanity check: 2(1)−(−1)−3=2+1−3=0. It works. …
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
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. …
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
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