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

For two distinct, non-vertical parallel lines, their gradients are equal: $m_1=m_2$. They point in the same direction and have the same steepness, so they never intersect.

Why Are the Gradients Equal?

The gradient $m$ measures the change in $y$ for each unit change in $x$:

$$m=\frac{\text{change in }y}{\text{change in }x}.$$

In the gradient-intercept form $y=mx+c$, $m$ determines the line's direction, while $c$ is the $y$-intercept. Therefore, distinct parallel lines have the same $m$ but different values of .

This can also be shown algebraically. If $y=m_1x+c_1$ and $y=m_2x+c_2$ intersect, equating them gives . When , this equation gives one intersection point. To avoid any intersection, distinct non-vertical lines therefore require and .

For example, consider

$$y=3x+2 \quad \text{and} \quad y=3x-5.$$

Both lines have gradient $3$, but their different intercepts mean they are distinct and parallel.

The vertical-line case must be handled separately. Lines $x=a$ and $x=b$, where $a\neq b$, are parallel, but their gradients are undefined because their horizontal change is zero.

Finding a Parallel Line

Suppose a line is parallel to $y=-2x+7$ and passes through $(3,4)$. Its gradient must be $-2$. Using point-gradient form,

$$y-4=-2(x-3),$$

so

$$y=-2x+10.$$

A common misconception is that equal gradients always indicate two distinct parallel lines. If both the gradient and intercept are equal, the equations represent the same line.

Exam Technique

This is IB Mathematics: Analysis and Approaches, SL 2.1, and it may appear on Paper 1 or Paper 2. For find, state $m_1=m_2$, substitute into $y-y_1=m(x-x_1)$, and simplify into the requested form. Show these stages because IB awards marks for method as well as the final equation.