# Parallel Straight Lines have Same Slope

Jump to navigation
Jump to search

## Theorem

Let $L_1$ and $L_2$ be straight lines in the plane.

Let $L_1$ and $L_2$ have slopes of $m_1$ and $m_2$ respectively.

Then $L_1$ and $L_2$ are parallel if and only if $m_1 = m_2$.

## Proof

Let $L_1$ and $L_2$ be embedded in a cartesian plane, given by the equations:

\(\displaystyle L_1: \ \ \) | \(\displaystyle y\) | \(=\) | \(\displaystyle m_1 x + c_1\) | ||||||||||

\(\displaystyle L_2: \ \ \) | \(\displaystyle y\) | \(=\) | \(\displaystyle m_2 x + c_2\) |

Let $\phi_1$ and $\phi_2$ be the angles that $L_1$ and $L_2$ make with the $x$-axis respectively.

Then from Slope of Straight Line is Tangent of Angle with Horizontal:

\(\displaystyle \tan \psi_1\) | \(=\) | \(\displaystyle m_1\) | |||||||||||

\(\displaystyle \tan \psi_2\) | \(=\) | \(\displaystyle m_2\) |

### Necessary Condition

Let $m_1 = m_2$.

Then:

- $\tan \psi_1 = \tan \psi_2$

and so:

- $\psi_1 = \psi_2 + n \pi$

The multiple of $\pi$ makes no difference.

Thus from Equal Corresponding Angles implies Parallel Lines, $L_1$ and $L_2$ are parallel.

$\Box$

### Sufficient Condition

Suppose $L_1 \parallel L_2$.

Then:

\(\displaystyle \phi_1\) | \(=\) | \(\displaystyle \phi_2\) | Parallelism implies Equal Corresponding Angles | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \tan \phi_1\) | \(=\) | \(\displaystyle \tan \phi_2\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle m_1\) | \(=\) | \(\displaystyle m_2\) |

$\blacksquare$

## Sources

- 1968: Murray R. Spiegel:
*Mathematical Handbook of Formulas and Tables*... (previous) ... (next): $\S 10$: Formulas from Plane Analytic Geometry: $10.9$: Angle $\psi$ between Two Lines having Slopes $m_1$ and $m_2$