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

Parallel lines have equal gradients. Perpendicular lines have gradients that are negative reciprocals, so for non-vertical lines their gradients satisfy $m_1m_2=-1$.

This is part of SL 2.1, Equations of a line in IB Mathematics: Applications and Interpretation.

The Relationship

The gradient $m$ measures the rate at which a line rises or falls as $x$ increases:

$$m=\frac{y_2-y_1}{x_2-x_1}.$$

Parallel lines point in the same direction, so they have the same rate of change. For example, $y=3x+2$ and $y=3x-5$ are parallel because both have gradient $3$, although their different intercepts keep them distinct.

For perpendicular lines, take the negative reciprocal of the first gradient. If $m_1=\frac{2}{5}$, then

$$m_2=-\frac{5}{2},$$

and checking gives

$$\frac{2}{5}\times\left(-\frac{5}{2}\right)=-1.$$

A common misconception is to change only the sign. For example, the perpendicular gradient to $4$ is not $-4$; it is $-\frac{1}{4}$.

The product rule has a special case: a vertical line has undefined gradient and is perpendicular to a horizontal line with gradient $0$. Two vertical lines are parallel, even though neither has a numerical gradient.

Worked Example

Suppose a line is perpendicular to $y=2x+7$ and passes through the point $(4,1)$. The known line has gradient $2$, so the required gradient is $-\frac{1}{2}$. Substitute this gradient and the point into :

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

After simplifying, the equation is $y=-\frac{1}{2}x+3$. Check that the gradient product is $2\times\left(-\frac{1}{2}\right)=-1$ and that substituting gives . These two checks confirm both required conditions.

To find a parallel line through the same point, retain gradient $2$:

$$y-1=2(x-4),$$

which simplifies to $y=2x-7$. Parallel lines must also have different intercepts if they are intended to be distinct; otherwise, the equations describe the same line.

Exam Technique

For an SL Paper 1-style question, identify the known gradient first. Write $m_2=m_1$ for a parallel line or $m_2=-\frac{1}{m_1}$ for a perpendicular line, then use the given point in to form the required equation.