# Condition for Straight Lines in Plane to be Parallel

## Theorem

Let $L: \alpha_1 x + \alpha_2 y = \beta$ be a straight line in $\R^2$.

Then the straight line $L'$ is parallel to $L$ iff there is a $\beta' \in \R^2$ such that:

- $L' = \set {\tuple {x, y} \in \R^2: \alpha_1 x + \alpha_2 y = \beta'}$

## Proof

### Necessary Condition

When $L' = L$, the claim is trivial.

Let $L' \ne L$ be described by the equation:

- $\alpha'_1 x + \alpha'_2 y = \beta'$

Without loss of generality, let $\alpha'_1 \ne 0$ (the case $\alpha'_2 \ne 0$ is similar).

Then for $\tuple {x, y} \in L'$ to hold, one needs:

\(\displaystyle \alpha'_1 x + \alpha'_2 y\) | \(=\) | \(\displaystyle \beta'\) | |||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle x\) | \(=\) | \(\displaystyle \frac {-\alpha'_2} {\alpha'_1} y + \frac {\beta'} {\alpha'_1}\) |

For $L'$ to be parallel to $L$, it is required that then $\tuple {x, y} \notin L$, that is:

\(\displaystyle \alpha_1 x + \alpha_2 y\) | \(\ne\) | \(\displaystyle \beta\) | |||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle \alpha_1 \paren {\frac {- \alpha'_2} {\alpha'_1} y + \frac {\beta'} {\alpha'_1} } + \alpha_2 y\) | \(\ne\) | \(\displaystyle \beta\) | ||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle \paren {\alpha_2 - \alpha_1 \frac {\alpha'_2} {\alpha'_1} } y + \alpha_1 \frac {\beta'} {\alpha'_1}\) | \(\ne\) | \(\displaystyle \beta\) | ||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle \paren {\alpha_2 - \alpha_1 \frac {\alpha'_2} {\alpha'_1} } y\) | \(\ne\) | \(\displaystyle \beta - \alpha_1 \frac {\beta'} {\alpha'_1}\) |

It follows that necessarily $\beta - \alpha_1 \frac {\beta'} {\alpha'_1} \ne 0$, or taking $y = 0$ would yield equality.

The only remaining way to obtain the desired inequality for all $y$ is that:

- $\alpha_2 - \alpha_1 \dfrac {\alpha'_2} {\alpha'_1} = 0$

One observes that now $\alpha_1 = 0 \implies \alpha_2 = 0$.

However, as $L: \alpha_1 x + \alpha_2 y = \beta$ is a straight line in $\R^2$, it cannot be that $\alpha_1 = \alpha_2 = 0$.

So $\alpha_1 \ne 0$, and one finds:

- $\alpha'_2 = \dfrac {\alpha'_1} {\alpha_1} \alpha_2$

Hence obtain:

\(\displaystyle \alpha'_1 x + \alpha'_2 y\) | \(=\) | \(\displaystyle \beta'\) | |||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle \frac {\alpha'_1} {\alpha_1} \paren {\alpha_1 x + \alpha_2 y}\) | \(=\) | \(\displaystyle \beta'\) | ||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle \alpha_1 x + \alpha_2 y\) | \(=\) | \(\displaystyle \beta' \frac {\alpha_1} {\alpha'_1}\) |

That is, $L'$ is described by an equation of the required form.

$\Box$

### Sufficient Condition

Let $L' \ne L$ be a straight line given by the equation:

- $\alpha_1 x + \alpha_2 y = \beta'$

Suppose we have a point $\mathbf x = \tuple {x, y} \in L \cap L'$.

Then, as $\mathbf x \in L$, it also satisfies:

- $\alpha_1 x + \alpha_2 y = \beta$

It follows that $\beta = \beta'$, so $L = L'$.

This contradiction shows that $L \cap L' = \O$, that is, $L$ and $L'$ are parallel.

The remaining case is when $L' = L$.

By definition, $L$ is parallel to itself.

The result follows.

$\blacksquare$

## Also see

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 28$