Gradient measures steepness: how much $y$ changes for every $1$ unit that $x$ increases. A positive gradient slopes upward left-to-right; a negative gradient slopes downward; a gradient of $0$ is a horizontal line.

Pick two clearly-plotted points on the line. Count the vertical rise and horizontal run between them:

A line rising $6$ squares over a run of $3$ squares has gradient $\frac{6}{3} = 2$.

Given $A(x_1, y_1)$ and $B(x_2, y_2)$:

This is the same rise-over-run idea, just done with subtraction instead of counting squares — it also works for points that aren't nicely placed on grid lines.

A horizontal line has no rise at all as $x$ changes, so its gradient is always $m = 0$. A vertical line, however, has no run — every point shares the same $x$-value, which means the formula would require dividing by $0$. A vertical line's gradient is therefore undefined, not zero; these two cases are easy to confuse but describe completely different situations.

Trace the line with your finger from left to right. If you're moving upward as you go, the gradient is positive; if you're moving downward, it's negative. This quick visual check is a useful way to sanity-check a calculated gradient before you commit to an answer — a line that clearly falls from left to right can never have come out with a positive $m$.

Every later coordinate-geometry topic depends on this single number. Parallel lines (E3.6) are recognised purely by having equal gradients; perpendicular lines (E3.7) are recognised by their gradients multiplying to $-1$; and the equation of any straight line (E3.5) can't be written down at all until its gradient has been found. Getting comfortable with rise-over-run now, including the sign conventions for uphill and downhill lines, removes the single biggest source of errors in everything that follows.

Find the gradient of the line joining $A(-3, 5)$ and $B(2, -5)$.

1. Let $(x_1,y_1) = (-3,5)$ and $(x_2,y_2)=(2,-5)$.
2. $m = \frac{-5 - 5}{2-(-3)} = \frac{-10}{5} = -2$
3. The gradient is $-2$ — the line falls $2$ units for every $1$ unit moved right.

A line passes through the grid points $(1, 2)$ and $(4, 8)$. Find its gradient.

1. Rise $= 8 - 2 = 6$. Run $= 4 - 1 = 3$.
2. $m = \frac{6}{3} = 2$
3. The gradient is $2$: the line rises $2$ units for every $1$ unit moved right, matching a clearly upward-sloping line.

Gradient is the foundation both relationships are built on: two lines are parallel exactly when their gradients are equal, and perpendicular exactly when their gradients multiply to give $-1$ (equivalently, one gradient is the negative reciprocal of the other). Calculating the gradient correctly here is therefore a prerequisite skill for E3.6 and E3.7, not just a standalone topic.