Parallel Straight Lines have Same Slope

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Theorem

Let $L_1$ and $L_2$ be straight lines in the plane.

Let $L_1$ and $L_2$ have slopes of $m_1$ and $m_2$ respectively.


Then $L_1$ and $L_2$ are parallel if and only if $m_1 = m_2$.


Proof

Let $L_1$ and $L_2$ be embedded in a cartesian plane, given by the equations:

\(\displaystyle L_1: \ \ \) \(\displaystyle y\) \(=\) \(\displaystyle m_1 x + c_1\)
\(\displaystyle L_2: \ \ \) \(\displaystyle y\) \(=\) \(\displaystyle m_2 x + c_2\)


Let $\phi_1$ and $\phi_2$ be the angles that $L_1$ and $L_2$ make with the $x$-axis respectively.

Then from Slope of Straight Line is Tangent of Angle with Horizontal:

\(\displaystyle \tan \psi_1\) \(=\) \(\displaystyle m_1\)
\(\displaystyle \tan \psi_2\) \(=\) \(\displaystyle m_2\)


Necessary Condition

Let $m_1 = m_2$.

Then:

$\tan \psi_1 = \tan \psi_2$

and so:

$\psi_1 = \psi_2 + n \pi$

The multiple of $\pi$ makes no difference.

Thus from Equal Corresponding Angles implies Parallel Lines, $L_1$ and $L_2$ are parallel.

$\Box$


Sufficient Condition

Suppose $L_1 \parallel L_2$.

Then:

\(\displaystyle \phi_1\) \(=\) \(\displaystyle \phi_2\) Parallelism implies Equal Corresponding Angles
\(\displaystyle \leadsto \ \ \) \(\displaystyle \tan \phi_1\) \(=\) \(\displaystyle \tan \phi_2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle m_1\) \(=\) \(\displaystyle m_2\)

$\blacksquare$


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