Condition for Straight Lines in Plane to be Parallel/Slope Form/Proof 2

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

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

\(\ds \leadsto \ \ \)

\(\ds \tan \phi_1\)

\(=\)

\(\ds \tan \phi_2\)

\(\ds \leadsto \ \ \)

\(\ds m_1\)

\(=\)

\(\ds m_2\)

$\blacksquare$