Perpendicular Lines

5.2.3 Perpendicular Lines

The coordinate axes themselves are perpendicular, and more generally any horizontal and any vertical line are perpendicular — in that special case one line has slope $0$ and the other has an undefined slope. For all other (non-vertical) perpendicular lines, there is a clean algebraic relationship between their slopes.

Theorem. Two non-vertical lines with slopes $m_1$ and $m_2$ are perpendicular to each other if and only if $m_1 \times m_2 = -1$.

Proof. Let $\alpha,\beta$ be the inclinations of the two lines, so $\tan\alpha = m_1$, $\tan\beta = m_2$, with $\alpha \ne 90^\circ$, $\beta \ne 90^\circ$ since the lines are non-vertical. If the lines are perpendicular, the geometry of the two possible configurations shows $\beta - \alpha = 90^\circ$ or $\alpha - \beta = 90^\circ$, i.e. $|\alpha - \beta| = 90^\circ$. Then

which is undefined, and this happens in the tangent-difference formula $\tan(\alpha-\beta) = \dfrac{\tan\alpha - \tan\beta}{1+\tan\alpha\tan\beta}$ exactly when the denominator is zero, i.e. $1 + \tan\alpha\tan\beta = 0$, i.e.

Conversely if $m_1 m_2 = -1$ the same identity forces $|\alpha - \beta| = 90^\circ$, so the lines are perpendicular. $\blacksquare$

Worked Example 1. Show $AB \perp BC$ for $A(1,2)$, $B(2,4)$, $C(0,5)$. Slope of $AB$, $m_1 = \dfrac{4-2}{2-1} = 2$. Slope of $BC$, $m_2 = \dfrac{5-4}{0-2} = -\dfrac12$. Then $m_1 m_2 = 2 \times \left(-\dfrac12\right) = -1$, so $AB \perp BC$. …