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For $b_1$ and $b_2$ non-zero, consider the lines $l_1=\{(x,y) \in \mathbb{R}^2 | a_1x + b_1y + c_1=0\}$ and $l_2=\{(x,y) \in \mathbb{R}^2 | a_2x + b_2y + c_1=0\}$.

Assuming I only know Euclid's postulates and the definition that lines are parallel if they do not intersect.

Start wearing purple
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flamingohats
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  • The intersection $l_1\cap l_2$ is the solution set of a $2\times 2$ linear system. You can use the determinant of this system to argue. – Julien Oct 20 '13 at 15:45

3 Answers3

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The two lines are parallel iff their slopes are equal, whici mean

$$\begin{align*}\text{slope of}\;\;l_1&: -\frac{a_1}{b_1}\\ \text{slope of}\;\;l_2&: -\frac{a_2}{b_2}\end{align*}\;\;\;,\;\;\;a_i\neq 0$$

Thus, the lines are parallel iff

$$\frac{a_1}{b_1}=\frac{a_2}{b_2}\iff a_1b_2-a_2b_1=0$$

I'll leave it to you the easy cases when $\;a_1=0\;$ or $\;a_2=0\;$ ...

DonAntonio
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  • Sir, I think above result holds only when we allow that, " line parallel to itself ". But why, according to Wikipedia definition (lines in a plane which do not meet) of parallel lines, line is not parallel to itself? – Akash Patalwanshi Jun 27 '21 at 15:59
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from Crammers rule system of two linear equations does not have solution if its determinant is equal $0$ so $$\det\begin{pmatrix}a_1&b_1\\a_2&b_2\end{pmatrix}=a_1b_2-a_2b_1=0$$

Pedro
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Adi Dani
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Assuming you only know the Euclidian postulates you'll get nowhere: Euclid's lines are definitely not given as the solution set of a statement form. That's due to Descartes.

Michael Hoppe
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  • Whether we represent the line as a solution set of a linear equation or not, in $\mathbb{R}^2$ they still follow Euclid's postulates. – robjohn Oct 20 '13 at 22:14