## 1. Two lines equation

Here I work in homogeneous coordinates (see ProjectiveCoordinates.html ) in which lines are represented by linear forms
L(X)=mx+ny+pz=0, and conics are represented by symmetric quadratic forms q(X,X)=ax2+by2+cz2+2fyz+2gzx+2hxy = 0.
I denote by the same letter  q(X,X') the corresponding bilinear form:
q(X,X') = axx'+byy'+czz'+f(yz'+y'z)+g(zx'+z'x)+h(xy'+x'y).
The property to show is the following one [SalmonConics, p. 270]:

Let a line L(X)=0 intersect the conic q(X,X)=0 at two points {X1,X2}. Then the couple of lines {L1=X0X1, L2=X0X2}
joining a point X0 with these two points can be represented as a product of lines through the (degenerate)
(0)                                               L2(X)q(X0,X0) -2L(X)L(X0)q(X,X0) + L2(X0)q(X,X) = 0.

To prove this consider an arbitrary line through X0 and a point X, which in parametric form is represented by:
(1)                                                                 X' = kX0 + mX.
The intersection point of this with L(X')=0 satisfies:
(2)                                                                 kL(X0) + mL(X) = 0 => (k : m) = (-L(X) : L(X0)).
If this point is also on the conic then q(X') = 0 translates to:
(3)          0 = q(X',X') = q(-L(X)X0 + L(X0)X,-L(X)X0 + L(X0)X) = L2(X)q(X0,X0) -2L(X)L(X0)q(X0,X) + L2(X0)q(X,X),
which is the claimed equation.

This simple property some interesting applications discussed below.

## 2. Fregier point

If a right angle turns about a point X0 lying on the conic c: q(X,X)=0, then the chord intercepted by its
legs on the conic passes through a constant point Y0 lying on the normal at X0 (see Fregier.html ).

To prove this use the natural homogeneous coordinates resulting by homogenization of the cartesian
coordinates i.e. replacing the cartesian (x', y') by (x'=x/z, y'=y/z) (see ProjectiveCoordinates.html ). In this system
if a product of lines L(X)*L'(Y) = (mx+ny+pz)(m'x+n'y+p'z) consists of orthogonal lines, then, since the quantity:
(4)                                                                                   mm' + nn',
represents the inner product of the normals of the two lines, this must be zero. The inverse is also true and gives
a necessary and sufficient condition for a degenerate quadratic to represent the product of two orthogonal
lines (in the natural homogeneous coordinate system).
Apply this criterion to the previous configuration, for which X0 is on the conic, hence q(X0,X0)=0. By (0):
(5)                                                            -2L(X)q(X,X0) + L(X0)q(X,X) = 0,
is the degenerate quadratic giving the product of the lines L1*L2=0. The coefficients of x2 and y2 in this are
easily seen to be:
(6)                                                      -2(amx0 + bny0) + (mx0+ny0+pz0)(a+b) =0.
Thus the variable line  L(X) = mx+ny+pz passes through the point:
(7)                                                          Y0 =  (  (b-a)x0, (a-b)y0, (a+b)z0 ).

Corollary The Fregier point Y0 of X0 varies on a conic c' as X0 varies on the conic  of reference c.

In fact (7) shows that points Y0 are transforms of X0:  Y0 = F(X0), where the map F is a projectivity
represented in homogeneous coordinates through the diagonal matrix   diag( b-a, a-b, a+b ).

## 3. General turning right angle

If a right angle turns about a point X0 not lying on the conic c: q(X,X)=0, then the chord intercepted by its
legs on the conic envelopes a second conic c'.

In this case the line joining the intersection points {x,x'} of the legs of the angle with the conic has coefficients
(r,s,t) and satisfies the equation of the first section:
(rx+sy+tz)2q(x0,x0)-2(rx+sy+tz)(rx0+sy0+tz0)q(x0,x)+ (rx0+sy0+tz0)2q(x,x)=0.
Again the equation mm'+nn'=0 for the sum of the coefficients of the terms x2 and y2, as explained in the previous section
leads this time to a quadratic equation with respect to (r,s,t) with coefficients depending on the coordinates (x0, y0, z0)
and the coefficients (a,b,c,f,g,h) of the conic. Thus the line envelopes a conic.

## 4. The vice of calculations

To satisfy the curiosity I do the calculations of the coefficients of the conic satisfied by (r,s,t).
First the sum of the coefficients of the terms x2 and y2 is:

Equating the sum of the coefficients to zero we obtain the equation in (r,s,t):