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Many coordinate geometry questions ask for a line through the point where two given lines meet, plus one more condition: through the origin, parallel to a line, equal intercepts. You can solve for the point first, but when it has awkward fractions that wastes time. The family of lines L1 + λL2 = 0 skips that step. This post explains why it works, works five examples with checks, and shows the one case it misses.
Where is this in the syllabus?
The JEE (Advanced) 2026 brochure lists "Lines through the point of intersection of two given lines" and concurrency of lines. The JEE (Main) 2026 syllabus lists intersection of lines and "conditions for concurrence of three lines", where the same idea is quick. NCERT's Class 11 chapter on straight lines (reprint 2026-27) includes miscellaneous exercises asking for lines through the intersection of two lines. Check the documents for your exam year.
Why does L1 + λL2 = 0 work?
Let L1 = a1x + b1y + c1 and L2 = a2x + b2y + c2, meeting at P. At P both expressions are zero, so L1 + λL2 is zero there for every λ. The equation is first degree in x and y, so it is a line. Each λ gives a different line through P. As λ grows very large the line approaches L2 but never becomes it, so L2 itself is the one line through P the family misses.
Worked examples
Use L1: x + 2y − 5 = 0 and L2: 3x − y − 1 = 0. They meet at (1, 2), which we will use only to check answers. The family is:
(x + 2y − 5) + λ(3x − y − 1) = 0, or (1 + 3λ)x + (2 − λ)y − (5 + λ) = 0
1. Through the origin. Put x = y = 0: −(5 + λ) = 0, so λ = −5. The line is −14x + 7y = 0, that is y = 2x. Check: (1, 2) lies on it.
2. Parallel to 4x + y = 7. The slope must be −4: −(1 + 3λ)/(2 − λ) = −4, so 1 + 3λ = 8 − 4λ and λ = 1. The line is 4x + y − 6 = 0. Check: 4 + 2 − 6 = 0.
3. Equal intercepts. The x-intercept is (5 + λ)/(1 + 3λ) and the y-intercept is (5 + λ)/(2 − λ). They are equal when 1 + 3λ = 2 − λ, so λ = 1/4, giving (7/4)x + (7/4)y − 21/4 = 0, that is x + y = 3. Check: 1 + 2 = 3. Note that λ = −5 makes both intercepts zero (the line y = 2x); some answer keys include it, so read the question.
4. Perpendicular to x + 3y = 0. The required slope is 3. Setting −(1 + 3λ)/(2 − λ) = 3 gives −1 − 3λ = 6 − 3λ, so −1 = 6, with no solution. That is the signal to test L2: 3x − y − 1 = 0 has slope 3 and passes through (1, 2). So the answer is 3x − y − 1 = 0.
5. Concurrency. Do L1, L2 and 5x + 3y − 11 = 0 meet at one point? Try to match the family: (1 + 3λ)/(2 − λ) = 5/3 gives 3 + 9λ = 10 − 5λ, so λ = 1/2. Then the family line is (5/2)x + (3/2)y − 11/2 = 0, which is 5x + 3y − 11 = 0 multiplied by 1/2. So yes. Check: 5 + 6 − 11 = 0 at (1, 2).
How do you find the fixed point of a family?
The idea also runs backwards. A question may say that (1 + k)x + (2k − 1)y − 3k = 0 passes through a fixed point for all k. Collect the terms with k: (x − y) + k(x + 2y − 3) = 0. This is a family through the intersection of x − y = 0 and x + 2y − 3 = 0. Solving, x = y and 3x = 3, so the fixed point is (1, 1). Check with k: (1 + k) + (2k − 1) − 3k = 0 for every k.
Which mistakes cost marks?
Forgetting the missing line. No λ does not mean no answer; test L2.
Writing the family as L1 + L2 + λ = 0. Adding a constant shifts the line away from P.
Slope sign slips. For ax + by + c = 0 the slope is −a/b.
Dropping the zero-intercept case in equal-intercept questions, or adding it when the question rules it out.
Not checking. Substituting the meeting point into your answer takes ten seconds.
A practice routine
Mixing question types helps you choose the method. In a study of 126 seventh-grade students over three months, Rohrer, Dedrick and Stershic (2015) found that interleaved practice, mixing different kinds of problems, gave higher test scores than blocked practice, both one day and 30 days later.
Two methods (weekly): solve three questions by finding the point first, then by the family, and compare the time taken.
Condition drill: for one pair of lines, find the family member for five different conditions.
Fixed points: rewrite three parameter families as L1 + kL2 = 0 and find the point.
Mix: add these to sets on forms, distances and angles of straight lines.
If you get stuck choosing which condition fixes λ, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer.
Today, find the line through the intersection of x + 2y − 5 = 0 and 3x − y − 1 = 0 that is perpendicular to 4x + y = 7, then check it at (1, 2).
