The Gradient Formula

An interval goes from $(1, 2)$ to $(5, 10)$. How far across is it, and how far up? Which subtraction did you do for each? Now write those two subtractions as a single fraction.

The rise is the change in $y$, which is $y_2 - y_1$. The run is the change in $x$, which is $x_2 - x_1$. Gradient was always rise over run, so the formula is nothing more than those two subtractions stacked.

You already define gradient as $\dfrac{\text{rise}}{\text{run}}$, and you already find the rise and run by counting squares. The only new step is naming those counts.

Take two points $(x_1, y_1)$ and $(x_2, y_2)$ and complete the right triangle beneath the interval, as in the diagram. Its horizontal leg stretches from $x_1$ to $x_2$, so its length is the difference $x_2 - x_1$. Its vertical leg stretches from $y_1$ to $y_2$, so its length is $y_2 - y_1$.

Substituting into $\dfrac{\text{rise}}{\text{run}}$ gives

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$

Nothing has changed except that the counting is now written down. That is what makes it usable when the coordinates are fractions, or when they are letters.

Call the points in either order and you get the same gradient. Take $(1, 2)$ and $(5, 10)$:

$\dfrac{10 - 2}{5 - 1} = \dfrac{8}{4} = 2$   and   $\dfrac{2 - 10}{1 - 5} = \dfrac{-8}{-4} = 2$

Both subtractions flipped, so both signs flipped, and the two negatives cancelled in the fraction.

The danger is flipping only one of them. Writing $\dfrac{10 - 2}{1 - 5} = \dfrac{8}{-4} = -2$ produces the negative of the right answer, and nothing in the arithmetic looks wrong. It is the single commonest error in this topic.

The habit that prevents it: label the points before you start. Decide which is $(x_1, y_1)$, write it down, and let both subtractions follow the labels.

The gradient carries two pieces of information at once, and they are independent.

The sign gives the direction. Positive means the interval rises as you move right; negative means it falls. This is a claim you can check against a sketch in a second, and you should.

The size gives the steepness. A gradient of $3$ is three times as steep as a gradient of $1$, and a gradient of $-3$ is just as steep as $3$, in the other direction. So when comparing steepness, compare $|m|$ and ignore the sign.

A gradient between $-1$ and $1$ describes an interval shallower than a diagonal. Beyond that range it is steeper than a diagonal. A gradient of $\tfrac{1}{2}$ and a gradient of $2$ are equally far from flat, in opposite senses, and that observation returns when perpendicular lines appear.

Two cases deserve their own treatment, and the formula explains both rather than breaking on them.

A horizontal interval joins points with the same y-coordinate, such as $(-5, 2)$ and $(3, 2)$. The rise is $2 - 2 = 0$, so $m = \dfrac{0}{8} = 0$. Zero divided by something is zero, and the gradient is $0$.

A vertical interval joins points with the same x-coordinate, such as $(4, -1)$ and $(4, 9)$. The run is $4 - 4 = 0$, so the formula asks for $\dfrac{10}{0}$, which has no value. The gradient is undefined.

Undefined is not the same as zero, and neither is it infinity. It means the question has no numerical answer, because there is no run to divide by. Saying a vertical line has gradient $0$ swaps the two cases and is worth being careful about.

Here is the formula doing something a picture cannot do reliably.

Three points $A$, $B$ and $C$ are collinear if they lie on a single straight line. Testing this by eye is unreliable, because points that are very nearly collinear look collinear.

The test is arithmetic. Find the gradient of $AB$ and the gradient of $BC$. If they are equal, then the two intervals have the same steepness and share the point $B$, so they must be parts of the same line. If the gradients differ, there is a bend at $B$ and the points are not collinear.

The shared point matters. Two separate intervals with equal gradients are merely parallel; it is the common point $B$ that forces them onto the same line rather than two parallel ones.

Five gradient items. Sketch each one roughly first and predict the sign before you calculate.

1. 1 Find the gradient of the interval joining $(0, 0)$ and $(4, 6)$.

   Show answer
   $\dfrac{6 - 0}{4 - 0}$, then simplify.$m = \tfrac{3}{2}$

2. 2 Find the gradient of the interval joining $(-3, 2)$ and $(5, 2)$.

   Show answer
   Equal y-coordinates, so the rise is zero.$m = 0$

3. 3 Find the gradient of the interval joining $(4, -1)$ and $(4, 9)$.

   Show answer
   Equal x-coordinates, so the run is zero and there is nothing to divide by.Undefined

4. 4 Find the gradient of the interval joining $(-2, -5)$ and $(1, 4)$.

   Show answer
   $\dfrac{4 - (-5)}{1 - (-2)} = \dfrac{9}{3}$$m = 3$

5. 5 Find the gradient of the interval joining $(6, 1)$ and $(2, 9)$.

   Show answer
   $\dfrac{9 - 1}{2 - 6} = \dfrac{8}{-4}$$m = -2$