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"Find k so that the three lines are concurrent" is quick if you know which method to use. Harder versions hide a catch: a zero determinant does not always mean the lines meet. This post shows three methods, when each is fastest, and how to handle the parallel case.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists "conditions for concurrence of three lines" in its straight-lines unit, and the JEE (Advanced) 2026 brochure lists "concurrency of lines". The NCERT Class 11 chapter Straight Lines (Reprint 2026-27) has a worked concurrency example (Example 11) and two exercises in its miscellaneous set (Questions 8 and 9). Check the documents for your exam year.
Method 1: solve two, substitute in the third
NCERT's definition: three lines are concurrent if the point of intersection of any two lies on the third.
Example (NCERT Miscellaneous Exercise, Q8). Find p so that 3x + y − 2 = 0, px + 2y − 3 = 0 and 2x − y − 3 = 0 meet at one point.
Add the first and third: 5x − 5 = 0, so x = 1. Then y = 2 − 3(1) = −1. The point is (1, −1).
Put it in the second line: p(1) + 2(−1) − 3 = 0, so p = 5.
Always solve the two lines without the unknown.
Method 2: the determinant
Write the lines as aᵢx + bᵢy + cᵢ = 0 for i = 1, 2, 3. If they meet at (x₀, y₀), then (x₀, y₀, 1) solves all three equations aᵢx + bᵢy + cᵢz = 0. A system like this has a solution other than (0, 0, 0) only when the determinant of its coefficients is zero. So the condition is that the 3 × 3 determinant with rows (a₁, b₁, c₁), (a₂, b₂, c₂) and (a₃, b₃, c₃) equals zero.
For the example above, with p = 5, the rows are (3, 1, −2), (5, 2, −3) and (2, −1, −3). Expanding along the first row: 3(−6 − 3) − 1(−15 + 6) + (−2)(−5 − 4) = −27 + 9 + 18 = 0. The two methods agree.
The determinant is fastest when the unknown appears in several lines. Our post on matrices and determinants for JEE covers expansion.
The catch: when a zero determinant misleads
The determinant is also zero when all three lines are parallel. Take x + y + 1 = 0, x + y + 2 = 0 and x + y + 3 = 0. The first two columns of the determinant are identical, so it is zero, but the lines never meet.
Worked example: a determinant with a repeated root
Find k so that kx + y + 1 = 0, x + ky + 1 = 0 and x + y + k = 0 are concurrent.
The determinant of rows (k, 1, 1), (1, k, 1), (1, 1, k) is k³ − 3k + 2 = (k − 1)²(k + 2).
So k = 1 or k = −2. Now check each.
k = 1: all three equations become x + y + 1 = 0. They are the same line, not three lines meeting at a point. Most questions reject this value.
k = −2: the lines are −2x + y + 1 = 0, x − 2y + 1 = 0 and x + y − 2 = 0. Each passes through (1, 1): −2 + 1 + 1 = 0, 1 − 2 + 1 = 0 and 1 + 1 − 2 = 0. So k = −2.
Skipping the check gives two answers where one is wrong.
Method 3: the coefficient trick for a family
Sometimes a question gives one condition on a, b and c and asks for the fixed point of all lines ax + by + c = 0.
Example. If 3a + 2b + 4c = 0, show that every line ax + by + c = 0 passes through a fixed point.
Divide the condition by 4: a(3/4) + b(1/2) + c = 0. This says the point (3/4, 1/2) satisfies ax + by + c = 0 for every such a, b and c. So all these lines are concurrent at (3/4, 1/2).
Our post on the family of lines through an intersection uses a related idea, L₁ + λL₂ = 0.
Worked example: NCERT's slope-intercept proof
NCERT Miscellaneous Exercise Q9: if y = m₁x + c₁, y = m₂x + c₂ and y = m₃x + c₃ are concurrent, show that m₁(c₂ − c₃) + m₂(c₃ − c₁) + m₃(c₁ − c₂) = 0.
Rewrite each as mᵢx − y + cᵢ = 0. The determinant with rows (mᵢ, −1, cᵢ) expands to exactly m₁(c₂ − c₃) + m₂(c₃ − c₁) + m₃(c₁ − c₂). Concurrency makes it zero.
Which mistakes cost marks?
Solving the pair that contains the unknown. Pick the two known lines.
Accepting every root of the determinant. Check for parallel or identical lines.
Sign errors in the constant term. Write every line as ax + by + c = 0 before building the determinant.
Forgetting the converse check. After finding k, substitute the point into all three lines.
A practice routine
Solve NCERT's Example 11 by both methods and confirm k = −2.
Make up three lines with a parameter in two of them, and solve with the determinant.
Find the fixed point of ax + by + c = 0 when a − 2b + 5c = 0.
If you set up a determinant and the algebra does not factor, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you can find where your expansion went wrong.
Today, find the values of a for which 2x + y − 1 = 0, ax + 3y − 3 = 0 and 3x + 2y − 2 = 0 are concurrent, and explain why the answer is unusual.
