Condition for Straight Lines in Plane to be Perpendicular/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 perpendicular to $L_2$ if and only if:

$\mu_1 = -\dfrac 1 {\mu_2}$


Proof 1

Let $\mu_1 = \tan \phi$.

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

\(\ds \mu_2\) \(=\) \(\ds \tan {\phi + \dfrac \pi 2}\) Definition of Perpendicular
\(\ds \) \(=\) \(\ds -\cot \phi\) Tangent of Angle plus Right Angle
\(\ds \) \(=\) \(\ds -\dfrac 1 {\tan \phi}\) Definition of Cotangent of Angle
\(\ds \) \(=\) \(\ds -\dfrac 1 {\mu_1}\) Definition of $\mu_1$

$\blacksquare$


Proof 2

This is a special case of Slope of Orthogonal Curves.

$\blacksquare$


Proof 3

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

From Angle between Straight Lines in Plane:

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

When $L_1$ and $L_2$ are perpendicular:

$\psi = \dfrac \pi 2$

by definition.


From Tangent of Right Angle $\tan \dfrac \pi 2$ is undefined.

This happens if and only if $1 + m_1 m_2 = 0$.

The result follows.

$\blacksquare$


Also presented as

This can also be presented in the elegant form:

$\mu_1 \mu_2 + 1 = 0$

or:

$\mu_1 \mu_2 = -1$


Sources

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