Gradient measures how steep a line is: how far it rises (or falls) for every step taken across.

A larger gradient (ignoring sign) means a steeper line; a gradient close to zero means a line that's almost flat. This single number completely describes the tilt of a straight line, regardless of where on the grid it happens to sit.

1. Choose two points on the line where it crosses grid intersections exactly.
2. Count the rise — squares moved up (positive) or down (negative) between the points.
3. Count the run — squares moved right between the points.
4. Divide: $m = \dfrac{\text{rise}}{\text{run}}$.

A line passes through $(0,1)$ and $(3,7)$: rise $=6$ (from $y=1$ to $y=7$), run $=3$ (from $x=0$ to $x=3$), so:

A line falls from $(-2,4)$ down to $(2,-4)$: rise $=-8$ (a fall, so negative), run $=4$, giving:

The negative sign matches the visual — the line runs downhill from left to right.

Gradient divides the rise by the run, but a perfectly vertical line has a run of exactly $0$ between any two of its points — and dividing by zero is undefined. This is why vertical lines are described separately, by the equation $x=k$, rather than being given a numerical gradient at all.

No — as long as both points sit exactly on grid intersections that the line passes through, any pair will give the same gradient, since a straight line rises and falls at a perfectly constant rate along its entire length. Picking points that are further apart (rather than right next to each other) often makes the rise and run easier to count accurately.

A line on a grid passes through $(-1,-3)$ and $(4,7)$, both exact grid intersections. Find its gradient, and state whether the line slopes up or down from left to right.

Solution

Rise: from $y=-3$ to $y=7$ is $10$ squares up. Run: from $x=-1$ to $x=4$ is $5$ squares right.

Since $m$ is positive, the line slopes upward from left to right.

The counting-squares method only works cleanly when both chosen points sit exactly where grid lines cross. If a line only passes exactly through grid intersections at certain points, always choose those over any point that merely looks close to a grid line — an estimated point introduces a small error that then affects the entire calculated gradient.