Condition for Straight Lines in Plane to be Parallel/Slope Form
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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 10$: Formulas from Plane Analytic Geometry: $10.9$: Angle $\psi$ between Two Lines having Slopes $m_1$ and $m_2$