ax+by+c=0 whoese coefficients (a,b,c) satisfy as + bt + c = 0. This is a homogeneous equation in (a,b,c) and setting arbitrary (a,b) we get c = -as-bt and the solutions:

(a,b,c) = a(1,0,-s) + b(0,1,-t).

This means that all lines of the pencil are expressible as a linear combination of the two lines:

x-s = 0 and y-t = 0.

Inversely given two lines f(x,y) = ax+by+c =0 , g(x,y) = a'x+b'y+c' = 0, every line of the form:

h(x,y) = sf(x,y) + tg(x,y) = (sa+ta')x+(sb+tb')x+(sc+tc') = 0,

defines a line of the pencil P*, where P is the intersection point of the two lines f(x,y)=0 and g(x,y)=0.

A particular member h(x,y) = sf(x,y)+tg(x,y) of the pencil P* is determined by the single condition to pass through a point X

h(x

This defines (s, t) up to a multiplicative constant (s,t) = k(-g(x

Thus the ratio which is essential is equal to:

(s/t) = -g(x

The formula with the sines is correct if the normals of the lines point both inside or outside, otherwise it has to be multiplied by -1.

The ratio |CA|/|CB| = area(ACP)/area(CPB) = |PA|sin(CPA)/(|PB|sin(CPB)) and since by the previous section the ratio of sines is equal to k:

This has the consequence to give directly the cross ratio of four lines in terms of the factor k. If the lines are f(x,y) = 0, g(x,y) = 0, f(x,y)-kg(x,y)=0, f(x,y)-k'g(x,y)=0, defining on j(x,y)=0 points {A,B,C,C'}, then the cross ratio (A,B,C,C') is:

Last formula for the cross ratio remains true even in the case in which lines f(x,y)=0, g(x,y)=0 are not in normal form. Because of the division the corresponding factors k

The discussion here is a sequel to the one started in Lines.html . It continues with the consideration of the quadratic equation defined by two lines contained in ConicsDegenerate.html .

Harmonic_Bundle.html

Lines.html

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