# pencil of lines

Let

${A}_{i}x+{B}_{i}y+{C}_{i}=0$ | (1) |

be equations of some lines. Use the short notations ${A}_{i}x+{B}_{i}y+{C}_{i}:={L}_{i}$.

If the lines ${L}_{1}=0$ and ${L}_{2}=0$ have an intersection^{} point $P$, then, by the parent entry (http://planetmath.org/LineThroughAnIntersectionPoint), the equation

${k}_{1}{L}_{1}+{k}_{2}{L}_{2}=0$ | (2) |

with various real values of ${k}_{1}$ and ${k}_{2}$ can any line passing through the point $P$; this set of lines is called a pencil of lines.

Theorem. A necessary and sufficient condition in to three lines

$${L}_{1}=0,{L}_{2}=0,{L}_{3}=0$$ |

pass through a same point, is that the determinant^{} formed by the coefficients of their equations (1) vanishes:

$$\left|\begin{array}{ccc}\hfill {A}_{1}\hfill & \hfill {B}_{1}\hfill & \hfill {C}_{1}\hfill \\ \hfill {A}_{2}\hfill & \hfill {B}_{2}\hfill & \hfill {C}_{2}\hfill \\ \hfill {A}_{3}\hfill & \hfill {B}_{3}\hfill & \hfill {C}_{3}\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill {A}_{1}\hfill & \hfill {A}_{2}\hfill & \hfill {A}_{3}\hfill \\ \hfill {B}_{1}\hfill & \hfill {B}_{2}\hfill & \hfill {B}_{3}\hfill \\ \hfill {C}_{1}\hfill & \hfill {C}_{2}\hfill & \hfill {C}_{3}\hfill \end{array}\right|=0.$$ |

Proof. If the line ${L}_{3}=0$ belongs to the fan of lines determined by the lines ${L}_{1}=0$ and ${L}_{2}=0$, i.e. all the three lines have a common point, there must be the identity

$${L}_{3}\equiv {L}_{1}+{L}_{2},$$ |

i.e. there exist three real numbers ${k}_{1}$, ${k}_{2}$, ${k}_{3}$, which are not all zeroes, such that the equation

${k}_{1}{L}_{1}+{k}_{2}{L}_{2}+{k}_{3}{L}_{3}\equiv 0$ | (3) |

is satisfied identically by all real values of $x$ and $y$.
This means that the group of homogeneous^{} linear equations

$\{\begin{array}{cc}{k}_{1}{A}_{1}+{k}_{2}{A}_{2}+{k}_{3}{A}_{3}=0,\hfill & \\ {k}_{1}{B}_{1}+{k}_{2}{B}_{2}+{k}_{3}{B}_{3}=0,\hfill & \\ {k}_{1}{C}_{1}+{k}_{2}{C}_{2}+{k}_{3}{C}_{3}=0\hfill & \end{array}$ |

has nontrivial solutions ${k}_{1},{k}_{2},{k}_{3}$. By linear algebra, it follows that the determinant of this group of equations has to vanish.

Suppose conversely that the determinant vanishes. This implies that the above group of equations has a nontrivial solution ${k}_{1},{k}_{2},{k}_{3}$. Thus we can write the identic equation (3). Let e.g. ${k}_{1}\ne 0$. Solving (3) for ${L}_{1}$ yields

$${L}_{1}\equiv -\frac{{k}_{2}{L}_{2}+{k}_{3}{L}_{3}}{{k}_{1}},$$ |

which shows that the line ${L}_{1}=0$ belongs to the fan determined by the lines ${L}_{2}=0$ and ${L}_{3}=0$; so the lines pass through a common point.

## References

- 1 Lauri Pimiä: Analyyttinen geometria. Werner Söderström Osakeyhtiö, Porvoo and Helsinki (1958).

Title | pencil of lines |
---|---|

Canonical name | PencilOfLines |

Date of creation | 2013-03-22 18:09:03 |

Last modified on | 2013-03-22 18:09:03 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 8 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 51N20 |

Related topic | LineInThePlane |

Related topic | Determinant2 |

Related topic | HomogeneousLinearProblem |

Related topic | Pencil2 |