## 1. Polar of point with respect to two lines

Given two lines {L1, L2} and a point P not lying on them, consider all secants L trhough P intersecting the lines respectively at points {Q1, Q2}. The geometric locus of the harmonic conjugate P' of P with respect to {Q1, Q2} is another line LP called the polar of P with respect to {L1, L2}.

The easiest way to prove this is to express the cross ratio (Q1,Q2,P,P') through the angles of the lines at Q. It is then (see CrossRatioLines.html )

Hence if this cross ratio is -1 for a line through P it is also -1 for all lines through P. Line QP' is called also the harmonic conjugate of line QP with respect to {L1, L2}.

## 2. The equation of the polar

Assume that the two lines {L1, L2} are given by equations f1(X)=0 and f2(X)=0. Then the polar of P with respect to {L1,L2} is given by
f2(P)f1(X) + f1(P)f2(X) = 0.
In fact, this is a line through the intersection point Q of {L1, L2} and consequently belongs to the pencil of lines
sf1(X) + tf2(x) = 0.
To find it notice that line PQ belongs also to that pencil, hence satisfies
sf1(P) + tf2(P) = 0,
and consequently
(s,t) = k(-f2(P), f1(P)).
Thus, line PQ has the equation
-f2(P)f1(X) + f1(P)f2(X) = 0.
The result follows by applying the previous relation of the sines and the fact that the ratio of the distances of P from the two lines is proportional to the ratio of sines:

## 3. The conjugate reflection

With the polar of a point P with respect to two lines {L1, L2} we have the possibility to define a naturally to this configuration related affine transformation. In fact, if LP denotes the polar of point P, then define the affine reflection RP by
RP = (PQ, LP).
The symbol refers to the above figure and defines the affine reflection (see Affine_Reflexion_Basics.html ) with axes of fixed points the line PQ and conjugate direction that of line LP. By definition this means that for every point X its image X' under the transformation has XX' parallel to LP and the middle of XX' is on line PQ.

## 4. Expression in coordinates

Assume that the line-equation is given in the form
f(X) =  (X,a) + c = 0,
where (X,Y)=x1y1+x2y2 is the usual inner product and c a constant. Assume further that lines {PQ, LP} are respectively given by two equations  {f1(X)=(X,a1)+c1=0  ,  f2(X)= (X,a2)+c2 = 0}. Then it is easily seen that the reflection is represented by the equation

Here JX denotes the π/2-rotated of vector X. Thus if X=(x1,x2), then JX = (-x2, x1). To express the transformation in terms of the original lines assume that {L1, L2} are given by equations { g1(X) = (b1,X)+d1=0 ,   g2(X)=2,X>+d2=0 }. Introducing the constants {s = g1(P), t = g2(P)} we saw that the axis PQ is given by
(a1,X)+c1 = -t((b1,X)+d1) + s((b2,X)+d2) = 0,
and the conjugate direction
(a2,X)+c2 = t((b1,X)+d1) + s((b2,X)+d2) = 0.
Hence
a1 = -tb1 + sb2, c1 = -td1+st2,
a2 =  tb1 + sb2, c1 =   td1+st2.
From these follows that
(a1, Ja2) = 2st(b2,Jb1)
and finally

## 5. The conjugate polygon

Given a polygon p=P0...Pn-1 with n sides a0=P0P1, a1=P1P2, ... etc. (indices exceeding n are taken modulo n) and a point Q, not on its side-lines, we can construct another polygon q=Q0...Qn1 with sides b0=Q0Q1, b1=Q1Q2, ... etc. by taking the polars of Q with respect to the side-pairs (ai, ai+1) of p. The resulting polygon circumscribes p and is called the conjugate of p with respect to Q.

### See Also

CrossRatioLines.html
Affine_Reflexion_Basics.html

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