Condition for Straight Lines in Plane to be Parallel/Slope Form

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Theorem

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

Let the slope of $L_1$ and $L_2$ be $\mu_1$ and $\mu_2$ respectively.


Then $L_1$ is parallel to $L_2$ if and only if:

$\mu_1 = \mu_2$


Proof 1

Let $L_1$ and $L_2$ be described by the general equation:

\(\ds L_1: \, \) \(\ds l_1 x + m_1 y + n_1\) \(=\) \(\ds 0\)
\(\ds L_2: \, \) \(\ds l_2 x + m_2 y + n_2\) \(=\) \(\ds 0\)


Then:

the slope of $L_1$ is $\mu_1 = -\dfrac {l_1} {m_1}$
the slope of $L_2$ is $\mu_2 = -\dfrac {l_2} {m_2}$.

From Condition for Straight Lines in Plane to be Parallel: General Equation:

$L_1$ and $L_2$ are parallel if and only if $L_2$ is given by the equation:
$m_1 x + m_2 y = n'$
for some $n'$.

But then the slope of $L_2$ is $-\dfrac {l_1} {m_1}$.

That is:

$-\dfrac {l_1} {m_1} = -\dfrac {l_2} {m_2}$

and the result follows.

$\blacksquare$


Proof 2

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

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


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

Then by definition of slope of a straight line:

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


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:

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

$\blacksquare$


Proof 3

Let $\psi$ be the angle between $L_1$ and $L_2$

When $L_1$ and $L_2$ are parallel:

$\psi = 0$

by definition.

From Angle between Straight Lines in Plane:

$\psi = \arctan \dfrac {m_1 - m_2} {1 + m_1 m_2}$

The result follows immediately.

$\blacksquare$


Sources

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