Angle Between Two Straight Lines described by Homogeneous Quadratic Equation

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\LL_1$ and $\LL_2$ represent $2$ straight lines in the Cartesian plane which are represented by a homogeneous quadratic equation $E$ in two variables:

$E: \quad a x^2 + 2 h x y + b y^2$


Then the angle $\psi$ between $\LL_1$ and $\LL_2$ is given by:

$\tan \psi = \dfrac {2 \sqrt {h^2 - a b} } {a + b}$


Proof

Let us rewrite $E$ as follows:

$b y^2 + 2 h x y + a x^2 = b \paren {y - \mu_1 x} \paren {y - \mu_2 x}$

Thus from Homogeneous Quadratic Equation represents Two Straight Lines through Origin:

$\LL_1$ and $\LL_2$ are represented by the equations $y = \mu_1 x$ and $y = \mu_2 x$ respectively.


From Sum of Roots of Quadratic Equation:

$\mu_1 + \mu_2 = -\dfrac {2 h} b$

From Product of Roots of Quadratic Equation:

$\mu_1 \mu_2 = \dfrac a b$


From Angle between Straight Lines in Plane:

$\tan \psi = \dfrac {\mu_1 - \mu_2} {1 + \mu_1 \mu_2}$

We have that:

\(\ds \paren {\mu_1 - \mu_2}^2\) \(=\) \(\ds \paren {\mu_1 + \mu_2}^2 - 4 \mu_1 \mu_2\)
\(\ds \) \(=\) \(\ds \paren {-\dfrac {2 h} b}^2 - 4 \dfrac a b\)
\(\ds \) \(=\) \(\ds \dfrac {4 \paren {h^2 - a b} } {b^2}\)
\(\ds \leadsto \ \ \) \(\ds \tan \psi\) \(=\) \(\ds \dfrac {\sqrt {4 \paren {h^2 - a b} / b^2} } {1 + a / b}\)
\(\ds \) \(=\) \(\ds \dfrac {2 \sqrt {\paren {h^2 - a b} } } {a + b}\) multiplying top and bottom by $b$ and simplifying

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Angle_Between_Two_Straight_Lines_described_by_Homogeneous_Quadratic_Equation&oldid=513289"