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}.
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:
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.
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
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.