Family of Lines and Concurrency: Explained with Examples

A family of lines is a collection of lines sharing a common property, described by one equation containing a parameter. The most used family is L1 + kL2 = 0, which represents every line passing through the intersection point of the two given lines L1 = 0 and L2 = 0 (except L2 = 0 itself). Three or more lines are concurrent if they all pass through one common point.

This idea matters because it converts "find a line satisfying some condition" problems into solving for a single parameter k instead of two unknowns m and c. Concurrency is central to triangle geometry — the three altitudes, the three medians and the three angle bisectors of any triangle are each concurrent — and determinant tests for concurrency appear in almost every coordinate-geometry exam.

Key Concepts

• Family through intersection: L1 + kL2 = 0, where L1 = a1x + b1y + c1 and L2 = a2x + b2y + c2. For every real k this is a line through the intersection point of L1 = 0 and L2 = 0.
• Excluded member: the family misses L2 = 0 itself; check it separately if the answer might be that line.
• Concurrency test: lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, a3x + b3y + c3 = 0 are concurrent if and only if the determinant of their coefficient matrix is zero.
• Parameter k: one extra condition (a point on the line, a slope, perpendicularity) fixes k.

Formulas

• Family: L1 + kL2 = 0, i.e. (a1x + b1y + c1) + k(a2x + b2y + c2) = 0.
• Concurrency determinant: | a1 b1 c1; a2 b2 c2; a3 b3 c3 | = 0.
• Assumption: L1 = 0 and L2 = 0 must not be parallel; otherwise there is no single intersection point and the family is a set of parallel lines.
• Units: none — k is a dimensionless real parameter.

Worked Example

Find the line through the intersection of x + y − 2 = 0 and 2x − y + 1 = 0 that also passes through (3, 1).

1. Write the family: (x + y − 2) + k(2x − y + 1) = 0.
2. Impose the extra point (3, 1): (3 + 1 − 2) + k(6 − 1 + 1) = 0, i.e. 2 + 6k = 0, so k = −1/3.
3. Substitute back: (x + y − 2) − (1/3)(2x − y + 1) = 0.
4. Multiply by 3: 3x + 3y − 6 − 2x + y − 1 = 0, giving x + 4y − 7 = 0.
5. Verify: (3, 1) gives 3 + 4 − 7 = 0. The intersection of the two given lines is (1/3, 5/3); substituting gives 1/3 + 20/3 − 7 = 7 − 7 = 0. Correct.

Common Misconceptions

• "L1 + kL2 = 0 gives every line through the point." It misses L2 = 0; if the required line happens to be L2, k cannot produce it.
• "Concurrent means the determinant of slopes is zero." The determinant test uses the full coefficients a, b, c of all three lines, not slopes.
• "Two lines through a point need slope–intercept form." The family method avoids dividing by slopes and handles vertical lines automatically.

Real-World Applications

• Computer graphics: families of lines through a vanishing point model perspective projection of parallel lines in 3D scenes.
• Structural design: concurrent force members meeting at a joint simplify truss analysis in civil engineering.
• Navigation: lines of position from two beacons intersect at the vessel's fix; a family through that fix tests candidate headings.
• Camera calibration: pencil-of-lines geometry is used to find the centre of projection from image measurements.

In short: L1 + kL2 = 0 packages infinitely many lines through one point into a single-parameter equation. Use one extra condition to fix k, remember the L2 exception, and test concurrency of three lines with the zero-determinant rule.