# CHASLES' THEOREM

#### By Hubert Shutrick

### 1. Pencils of Lines

Let *f*_{s} = *a*_{s}*x* + *b*_{s}*y* + *c* and *f*_{t} = *a*_{t}*x* + *b*_{t}*y* + *c* where *f*_{s} = 0*f*_{t} = 0*s*, *t* such that *s*, *t*) ≠ (0, 0)

*sf*_{s} + *tf*_{t} = 0,

and these lines form a *pencil of lines* with *base* *f*_{s}, *f*_{t}. For
*r* ≠ 0,*rs*, *rt*)*s*, *t*)*homogeneous*. If the lines *f*_{s} = 0*f*_{t} = 0*f* = 0

are linearly dependent using the common point, and, since the first two rows are independent, the last row must be a linear combination of the first two. Hence, *f* = 0*f*_{s} = 0*f*_{t} = 0*f* = 0

### 2. Change of Base

If

f_{u} | = s_{u} f_{s} + t_{u} f_{t} | |

f_{v} | = s_{v} f_{s} + t_{v} f_{t} | |

give two distinct lines of the pencil, then *f*_{u}, *f*_{v} can be a new base.

uf_{u} + vf_{v} | = u(s_{u} f_{s} + t_{u} f_{t}) + v(s_{v} f_{s} + t_{v} f_{t}) | |

= (us_{u} + vs_{v})f_{s} + (ut_{u} + vt_{v})f_{t}. | ||

The transition from (*u*, *v*) to *s*, *t*)

It is important to note that the base is determined by *f*_{s}, *f*_{t} but not by the lines *f*_{s} = 0*f*_{t} = 0*s*_{u} ≠ *t*_{v}*f*_{u} = *s*_{u} *f*_{s},*f*_{v} = *t*_{v} *f*_{t}*s*_{u} *f*_{s} + *t*_{v} *f*_{t} = 0

### 3. Cross-ratio

In terms of the homogeneous parameters for the pencil, the convenient definition of cross-ratio is

It agrees with the one parameter version if the parameter is replaced by *s*/*t*.

Its importance in this context is that it is depends only on the four lines and not on the choice of homogeneous parameters. This becomes obvious using that the determinant of the product of two matrices is the product of the determinants:

If three different points are given along with the cross-ratio *k* of them with a forth point, then the forth point is uniquely determined. To prove this, choose the base so that the points have parameters *s*, *t*).

### 4. Projective Plane

The *projective plane* is obtained from the ordinary Euclidean plane by adjoining an extra point for each pencil of parallel lines. It is the point 'at infinity' where the lines of the pencil meet. To accommodate the new points it is convenient to introduce *homogeneous coordinates*. This is done by replacing the usual *x*, *y**x*/*z*, *y*/*z*.*x*, *y*, *z*) ≠ (0, 0, 0)*r* ≠ 0,*rx*, *ry*, *rz*)

A line in the plane is given by an equation *ax* + *by* + *cz* = 0,*z* = 0

meet at (*b*, -*a*, 0) as required.

The line joining points *x*_{s}, *y*_{s}, *z*_{s})*x*_{t}, *y*_{t}, *z*_{t})*s*, *t*

*s*(*x*_{s}, *y*_{s}, *z*_{s}) + *t* (*x*_{t}, *y*_{t}, *z*_{t})

is on the line for each *s*, *t*) ≠ (0, 0).*x*_{0}, *y*_{0}, *z*_{0})*s*, *t*

This can be rewritten

The parameters for the line are the parameters for the pencil of lines through *x*_{0}, *y*_{0}, *z*_{0}).

### 5. Change of Coordinates

The coordinate vectors *X* = (1, 0, 0), *Y* = (0, 1, 0),*Z* = (0, 0, 1)*base* for the coordinate system in the sense that *x*, *y*, *z*) = *xX* + *yY* + *zZ*.

X' | = (x_{x}, y_{x}, z_{x}) | |

Y' | = (x_{y}, y_{y}, z_{y}) | |

Z' | = (x_{z}, y_{z}, z_{z}) | |

are not collinear, then they can form a new base:

(*x*, *y*, *z*) = *x*'*X*' + *y*'*Y*' + *z*'*Z*'.

In matrix form

The matrix is invertible since the points are not collinear.

Note that the points defined by *X*, *Y*, *Z*

X' | = x_{x} X | |

Y' | = y_{y} Y | |

Z' | = z_{z} Z | |

defines a new base with the same base points if *x*_{x}, *y*_{y}, *z*_{z}*x*_{x}, *y*_{y}, *z*_{z})

### 6. Chasles' Theorem

The theorem states that,

for four points on a non-degenerate conic, the cross-ratio of the pencil of four lines from a fifth point of the conic to the given points does not depend on the choice of the fifth point.

To prove this choose a coordinate system such that *X* and *Y* are points of the conic, *Z* is the intersection of the tangents from *X* and *Y* and the

*ax*² + *by*² + *cz*² + *fyz* + *gxz* + *hxy* = 0.

Since the line *y* = 0 is tangent at *a* = *g* = 0*z* = 0.*b* = *f* = 0.

*xy* = *z*².

Consider the pencil of lines through *X* with base *y*, -*z*.*s*, *t**sy* = *tz**s*², *t*², *st*).*s*, *t**Y* with base *z*, -*x*.*sz* = *tx**s*², *t*², *st*);*X* and *Y* to four points of the conic must be the same. Since *X* and *Y* were arbitrary points of the conic, the theorem is true.

(Chasles' theorem leads to an elegant projective proof of Pascal's Theorem. Also, there is another proof of Chasles' Theorem and a third, the simplest of all.)

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