Condition for Perpendicular Lines

Two non-vertical lines are perpendicular if and only if the product of their slopes is −1.

Let two lines have inclinations $\theta_1$ and $\theta_2$. If they are perpendicular:

$$\theta_2 = \theta_1 + 90°$$

Then:

$$m_2 = \tan\theta_2 = \tan(\theta_1 + 90°) = -\cot\theta_1 = -\frac{1}{\tan\theta_1} = -\frac{1}{m_1}$$

Therefore:

$$\boxed{m_1 m_2 = -1}$$

Geometric interpretation

If line $L_1$ has slope $m$, a vector along it is $(1, m)$. A vector along a line perpendicular to $L_1$ is $(-m, 1)$ (rotate 90°). The slope of the perpendicular is $\frac{1}{-m} = -\frac{1}{m}$, giving $m \cdot (-\frac{1}{m}) = -1$.

Edge cases

If one line is horizontal ($m_1 = 0$), the perpendicular is vertical (undefined slope) — handle separately.

If one line is vertical (undefined slope), the perpendicular is horizontal ($m = 0$) — handle separately.

> > $$ > \vec{d_1} \cdot \vec{d_2} = 0 > $$ > > Computing: $(1)(1) + (m_1)(m_2) = 1 + m_1 m_2 = 0$, which gives $m_1 m_2 = -1$. > > The dot product condition $\vec{d_1} \cdot \vec{d_2} = 0$ is the universal test — it works in 2D, 3D, and any dimension, and handles vertical/horizontal lines without special cases (use direction vectors $(0,1)$ and $(1,0)$ respectively).