What's the Relationship Between the Gradients of Perpendicular Lines?

For two non-vertical perpendicular lines, their gradients are negative reciprocals: if one gradient is $m_1$, the other is $m_2=-\frac{1}{m_1}$. Equivalently, $m_1m_2=-1$. A horizontal line and a vertical line are also perpendicular, although the vertical line has undefined gradient.

This relationship is in IB Mathematics: Analysis and Approaches, SL 2.1: Equations of straight lines, parallel and perpendicular.

Why Are the Gradients Negative Reciprocals?

A gradient measures vertical change divided by horizontal change: $m=\frac{\text{rise}}{\text{run}}$. Suppose a direction vector for a line is ((b,a)), giving gradient (a/b). Rotating this vector through $90^\circ$ gives ((-a,b)) or ((a,-b)). Therefore, the new gradient is

$$m_{\perp}=-\frac{b}{a}=-\frac{1}{m}.$$

Multiplying the two gradients gives $-1$, which is the algebraic test for perpendicular non-vertical lines. Geometrically, this rotation changes the direction by a right angle.

Worked Example

A line with gradient $\frac{2}{3}$ has perpendicular gradient $-\frac{3}{2}$. If the perpendicular line passes through , use :

$$y-1=-\frac{3}{2}(x-4).$$

Expanding gives $y=-\frac{3}{2}x+7$. This demonstrates both required steps: find the perpendicular gradient and then use the given point.

Important Cases

A common misconception is that perpendicular gradients are simply opposites. The perpendicular gradient to $2$ is not $-2$; it is $-\frac{1}{2}$. Another error is applying $m_1m_2=-1$ to a vertical line, whose gradient does not exist. In that special case, identify horizontal and vertical lines directly.

Exam Technique

In an IB response, the command term find requires relevant working. Write $m_{\perp}=-1/m$ before substituting into $y-y_1=m(x-x_1)$. If equations are in general form, rearrange them to identify each gradient, or compare direction vectors. Show the negative-reciprocal step explicitly to secure the method mark, and state the final equation clearly.

When asked to justify perpendicularity, calculate both gradients and show their product is $-1$, rather than relying only on the appearance of a graph.