# Angle Between Two Straight Lines described by Homogeneous Quadratic Equation

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## 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

- 1933: D.M.Y. Sommerville:
*Analytical Conics*(3rd ed.) ... (previous) ... (next): Chapter $\text {II}$. The Straight Line: $16$. Angle between the two straight lines represented by the homogeneous equation