# Condition for Straight Lines in Plane to be Perpendicular/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 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 $L_2$

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

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:

$\mu_1 \mu_2 = -1$