Condition for Pairs of Lines through Origin to be Harmonic Conjugates
Theorem
Consider $4$ lines $\LL_1$, $\LL_2$, $\LL_3$ and $\LL_4$ through the origin $O$ whose equations embedded in the Cartesian plane are as follows:
\(\ds (\LL_1): \ \ \) | \(\ds y\) | \(=\) | \(\ds \lambda x\) | |||||||||||
\(\ds (\LL_2): \ \ \) | \(\ds y\) | \(=\) | \(\ds \mu x\) | |||||||||||
\(\ds (\LL_3): \ \ \) | \(\ds y\) | \(=\) | \(\ds \lambda' x\) | |||||||||||
\(\ds (\LL_4): \ \ \) | \(\ds y\) | \(=\) | \(\ds \mu' x\) |
Then the pairs of lines $\set {\LL_1, \LL_2}$ and $\set {\LL_3, \LL_4}$ are harmonic conjugates if and only if:
- $\dfrac {\lambda - \lambda'} {\lambda - \mu'} / \dfrac {\mu - \lambda'} {\mu - \mu'} = -1$
Homogeneous Quadratic Equation Form
Consider the two homogeneous quadratic equations:
\(\text {(E1)}: \quad\) | \(\ds a_1 x^2 + 2 h_1 x y + b_1 y^2\) | \(=\) | \(\ds 0\) | |||||||||||
\(\text {(E2)}: \quad\) | \(\ds a_2 x^2 + 2 h_2 x y + b_2 y^2\) | \(=\) | \(\ds 0\) |
each representing two straight lines through the origin.
Then the two straight lines represented by $(\text E1)$ are harmonic conjugates of the two straight lines represented by $(\text E2)$ if and only if:
- $a_1 b_2 + a_2 b_1 - 2 h_1 h_2 = 0$
Proof
Let $\set {\LL_1, \LL_2}$ and $\set {\LL_3, \LL_4}$ be harmonic conjugates as asserted.
A straight line in the plane which does not pass through $O$ will either:
- intersect all four of $\LL_1$, $\LL_2$, $\LL_3$ and $\LL_4$
or:
- be parallel to one such straight line and intersect the other three.
Hence, let a straight line be drawn parallel to $\LL_1$ so as to intersect each of $\LL_2$, $\LL_3$, $\LL_4$ at $L'$, $M$ and $M'$.
It would cut $\LL_1$ at $L$, but $L$ is the point at infinity of $\LL_1$.
From Straight Line which cuts Harmonic Pencil forms Harmonic Range, the points of intersection define a harmonic range.
We have that $L'M'$ is parallel to $\LL_1$.
From Harmonic Range with Unity Ratio, the harmonic range defined by $L'M'$ is such that $M$ is the midpoint of $L'M'$.
$L'M'$ can be expressed in slope-intercept form as $y = \lambda x + c$ for some $c \in \R_{\ne 0}$.
Hence the abscissae of the points $L'$, $m$ and $M'$ are respectively:
- $\dfrac c {\lambda' - \lambda}$, $\dfrac c {\mu - \lambda}$, $\dfrac c {\mu' - \lambda}$
Hence the condition that the pairs of lines $\set {\LL_1, \LL_2}$ and $\set {\LL_3, \LL_4}$ are harmonic conjugates is:
- $\dfrac 2 {\mu - \lambda} = \dfrac 1 {\lambda' - \lambda} + \dfrac 1 {\mu' - \lambda}$
which is more usually expressed as:
- $\dfrac {\lambda - \lambda'} {\lambda - \mu'} / \dfrac {\mu - \lambda'} {\mu - \mu'} = -1$
$\blacksquare$
Also presented as
This result can also be presented as:
- $\dfrac {\paren {\lambda - \lambda'} \paren {\mu - \mu'} } {\paren {\lambda - \mu'} \paren {\mu - \lambda'} } = -1$
Sources
- 1933: D.M.Y. Sommerville: Analytical Conics (3rd ed.) ... (previous) ... (next): Chapter $\text {II}$. The Straight Line: $20$. Condition that the two pairs of lines through the origin, $y = \lambda x$, $y = \mu x$ and $y = \lambda' x$, $y = \mu' x$ should be apolar