Condition for Straight Lines in Plane to be Parallel/General Equation

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:


 * $L' = \left\{{ \left({x, y}\right) \in \R^2 : \alpha_1 x + \alpha_2 y = \beta\,' }\right\}$

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 $\left({x, y}\right) \in L'$ to hold, one needs:

For $L'$ to be parallel to $L$, it is required that then $\left({x, y}\right) \notin L$, i.e.:

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:

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

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 $\mathbf x = \left({x, y}\right) \in L \cap L'$.

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 $L \cap L' = \varnothing$, i.e., $L$ and $L'$ are parallel.

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

By definition, $L$ is parallel to itself.

The result follows.

Also See

 * Condition for Planes being Parallel