Condition for Straight Lines being 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:

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

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

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

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

	\(\displaystyle \)	\(\displaystyle \alpha_1 x + \alpha_2 y\)	\(\ne\)	\(\displaystyle \beta\)	\(\displaystyle \)
	\(\displaystyle \iff\)	\(\displaystyle \alpha_1 \left({ \frac {- \alpha'_2} {\alpha'_1} y + \frac {\beta'} {\alpha'_1} }\right) + \alpha_2 y\)	\(\ne\)	\(\displaystyle \beta\)	\(\displaystyle \)
	\(\displaystyle \iff\)	\(\displaystyle \left({\alpha_2 - \alpha_1 \frac {\alpha'_2} {\alpha'_1} }\right) y + \alpha_1 \frac {\beta'} {\alpha'_1}\)	\(\ne\)	\(\displaystyle \beta\)	\(\displaystyle \)
	\(\displaystyle \iff\)	\(\displaystyle \left({\alpha_2 - \alpha_1 \frac {\alpha'_2} {\alpha'_1} }\right) y\)	\(\ne\)	\(\displaystyle \beta - \alpha_1 \frac {\beta'} {\alpha'_1}\)	\(\displaystyle \)

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:

	\(\displaystyle \)	\(\displaystyle \alpha'_1 x + \alpha'_2 y\)	\(=\)	\(\displaystyle \beta'\)	\(\displaystyle \)
	\(\displaystyle \iff\)	\(\displaystyle \dfrac {\alpha'_1} {\alpha_1} \left({ \alpha_1 x + \alpha_2 y }\right)\)	\(=\)	\(\displaystyle \beta'\)	\(\displaystyle \)
	\(\displaystyle \iff\)	\(\displaystyle \alpha_1 x + \alpha_2 y\)	\(=\)	\(\displaystyle \beta' \dfrac {\alpha_1} {\alpha'_1}\)	\(\displaystyle \)

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

Then, as $\vec x \in L$, 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.

$\blacksquare$

Also See

• Condition for Planes being Parallel

Sources

• Seth Warner: Modern Algebra (1965): $\S 28$