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 $L$ if and only if 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:
\(\ds \alpha'_1 x + \alpha'_2 y\) | \(=\) | \(\ds \beta'\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(=\) | \(\ds \frac {-\alpha'_2} {\alpha'_1} y + \frac {\beta'} {\alpha'_1}\) |
For $L'$ to be parallel to $L$, it is required that then $\tuple {x, y} \notin L$, that is:
\(\ds \alpha_1 x + \alpha_2 y\) | \(\ne\) | \(\ds \beta\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \alpha_1 \paren {\frac {-\alpha'_2} {\alpha'_1} y + \frac {\beta'} {\alpha'_1} } + \alpha_2 y\) | \(\ne\) | \(\ds \beta\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \paren {\alpha_2 - \alpha_1 \frac {\alpha'_2} {\alpha'_1} } y + \alpha_1 \frac {\beta'} {\alpha'_1}\) | \(\ne\) | \(\ds \beta\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \paren {\alpha_2 - \alpha_1 \frac {\alpha'_2} {\alpha'_1} } y\) | \(\ne\) | \(\ds \beta - \alpha_1 \frac {\beta'} {\alpha'_1}\) |
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:
\(\ds \alpha'_1 x + \alpha'_2 y\) | \(=\) | \(\ds \beta'\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \frac {\alpha'_1} {\alpha_1} \paren {\alpha_1 x + \alpha_2 y}\) | \(=\) | \(\ds \beta'\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \alpha_1 x + \alpha_2 y\) | \(=\) | \(\ds \beta' \frac {\alpha_1} {\alpha'_1}\) |
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:
- $\alpha_1 x + \alpha_2 y = \beta'$
Aiming for a contradiction, 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
- 1933: D.M.Y. Sommerville: Analytical Conics (3rd ed.) ... (previous) ... (next): Chapter $\text {II}$. The Straight Line: $5$
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text V$: Vector Spaces: $\S 28$: Linear Transformations