Slope (gradient) — definition, formula, interpretation: Worked examples

This page is the practice arena for Slope (gradient). The parent note built the formula; here we hunt down every single case the formula can land in — every sign, every zero, every trap — so that when an exam throws one at you, you have already met its twin.

The scenario matrix

Before working problems, we list every class of case a slope question can belong to. Think of this as a checklist — by the end, every row has a worked example next to it.

#	Case class	What makes it special	Example that hits it
A	Positive slope (uphill →)	$Δ y$ and $Δ x$ same sign	Ex 1
B	Negative slope (downhill →)	$Δ y$ and $Δ x$ opposite signs	Ex 2
C	Order-independence	swap the two points, same answer	Ex 3
D	Zero slope (horizontal)	$Δ y = 0$ , run $\neq = 0$	Ex 4, part (D)
E	Undefined slope (vertical)	$Δ x = 0$ → divide by zero	Ex 4, part (E)
F	Fractional / shallow slope	$	m
G	Steepness comparison	which of two lines is steeper?	Ex 6
H	Slope → angle ( $m = tan θ$ ), incl. negative & steep	recover the inclination	Ex 7
I	Real-world word problem	slope carries units	Ex 8
J	Exam twist: solve for an unknown coordinate	slope given, find a missing $x$ or $y$	Ex 9
K	Collinearity via equal slopes	three points on one line?	Ex 10
L	Indeterminate case: same point twice	$Δ x = 0$ and $Δ y = 0$	Ex 11

Note on Ex 4: it is a single example with two sub-parts — part (D) is the horizontal (zero-slope) case and part (E) is the vertical (undefined-slope) case. That is why both rows D and E point to the same worked example.

We will now walk them in order. Signs and geometry first (A–E), then the subtler ones (F–L).

Signs and degenerate cases (A–E)

Figure — Slope (gradient) — definition, formula, interpretation

Figure 1 — The four flavours of slope. Four lines share one corner point. The pale-yellow line climbs to the right (positive slope, labelled $m > 0$ ); the pink line falls to the right (negative slope, $m < 0$ ); the blue horizontal line never rises (zero slope, $m = 0$ ); the white dashed vertical line has no run at all (undefined slope). Each example below is one of these four flavours — keep this picture in mind.

Worked example Example 4 — Cases D & E: horizontal and vertical

This one example has two parts: part (D) is the horizontal case (row D of the matrix), part (E) is the vertical case (row E). (D) Line through $(0, 3)$ and $(7, 3)$ . (E) Line through $(5, 1)$ and $(5, 9)$ .

Forecast: one of these has slope $0$ ; the other has no slope. Which is which — and why is "no slope" different from "zero slope"?

(D) Rise $= 3 - 3 = 0$ ; run $= 7 - 0 = 7$ . So $m = \frac{0}{7} = 0$ . Why this step? Zero divided by a non-zero number is genuinely $0$ — a flat line, no climbing.
(E) Rise $= 9 - 1 = 8$ ; run $= 5 - 5 = 0$ . So $m = \frac{8}{0} =$ undefined. Why this step? Division by zero has no meaning; the line rises with no sideways movement — a wall.

Verify: $\frac{0}{7} = 0$ ✓ (horizontal). For (E), you cannot answer "how much up per step right" when you can take no step right — so "undefined", never " $0$ ". These are opposite ends, not the same.

Slope size, angle, units and twists (F–L)

Worked example Example 7 — Case H: slope to angle of inclination (all signs)

(a) A line has slope $m = 1$ . (b) Another has slope $m = - 1$ . What angle $θ$ does each make with the positive $x$ -axis? (See Angle of Inclination.)

Forecast: rise equals run in (a) — the classic "diagonal". In (b) it falls at the same steepness. Guess both angles.

Set up the tool: in the right triangle of rise and run, $m = \frac{opposite}{adjacent} = tan θ$ . Why THIS tool (tangent, not sine)? Sine would need the hypotenuse; we only have opposite (rise) and adjacent (run). Tangent is the only ratio built from exactly those two — so it is the natural bridge from slope to angle.
Undo the tangent: $θ = arctan (m)$ . Why arctan? $arctan$ answers "which angle has this tangent?" — it undoes $tan$ .
(a) Evaluate: $arctan (1) = 45°$ .
(b) Evaluate: $arctan (- 1) = - 45°$ . Why this step? $arctan$ returns angles between $- 90°$ and $+ 90°$ , so a negative slope gives a negative angle — the line dips $45°$ below the positive $x$ -axis.
Convert to a standard inclination — with quadrant reasoning. The angle of inclination is conventionally measured $0°$ to $180°$ . For a downhill line we can realise slope $- 1$ as rise $= + 1$ , run $= - 1$ : we walk left ( $Δ x < 0$ ) and up ( $Δ y > 0$ ). A point that is up-and-to-the-left of the origin sits in the second quadrant (top-left). So the line's true inclination — measured anticlockwise from the positive $x$ -axis — must land between $90°$ and $180°$ , i.e. an obtuse angle. The raw $arctan (- 1) = - 45°$ points into the fourth quadrant (down-right); adding $180°$ rotates it into the correct second-quadrant direction: $- 45° + 180° = 135°$ . Why add exactly $180°$ ? Because $tan$ repeats every $180°$ ( $tan (θ) = tan (θ + 180°)$ ), the two directions $- 45°$ and $135°$ share the same tangent $- 1$ — they are the same line pointing opposite ways. We pick the one in $[0°, 180°)$ , which is $135°$ .

Verify: $tan 45° = 1$ ✓ and $tan 135° = - 1$ ✓ (both give back the correct slope). Note the pattern: $m > 0 \Rightarrow θ$ between $0°$ and $90°$ (first quadrant direction); $m < 0 \Rightarrow θ$ between $90°$ and $180°$ (second quadrant direction); $m = 0 \Rightarrow θ = 0°$ ; and a vertical line ( $m$ undefined) is the limiting $θ = 90°$ — the one inclination $arctan$ can never output because $tan 90°$ is itself undefined.

Worked example Example 8 — Case I: real-world word problem (with units!)

A wheelchair ramp rises $0.4 m$ over a horizontal distance of $5 m$ . Building code allows a maximum slope of $1 : 12$ ( $\approx 0.083$ ). Does this ramp pass?

Forecast: $0.4$ up over $5$ across — feels gentle. But is it under the legal limit?

Rise $= 0.4 m$ , run $= 5 m$ .
Slope $= \frac{0.4}{5} = 0.08$ (metres up per metre across — the units cancel to a pure ratio). Why this step? Slope here is a genuine gradient; because both are in metres, $m$ is dimensionless.
Compare: $0.08 < 0.0833 \dots$ , so the ramp is just under the limit. Why this step? $1 : 12$ means $1$ up per $12$ across, i.e. $1/12 = 0.0833 \dots$ ; a smaller number is a gentler ramp.

Verify: $1 : 12 = 1/12 = 0.0833 \dots$ ; our $0.08 = 1/12.5$ , i.e. $1 : 12.5$ , which is gentler than $1 : 12$ . Ramp passes. ✓ This is the Rate of Change idea: metres of height gained per metre travelled.

Worked example Example 10 — Case K: collinearity via equal slopes

Are $A = (1, 1)$ , $B = (3, 5)$ , $C = (6, 11)$ on one straight line?

Forecast: three points lie on one straight line exactly when every pair of them gives the same slope — because a single line has one slope everywhere. Guess before checking.

Slope $A B$ : $\frac{5 - 1}{3 - 1} = \frac{4}{2} = 2$ . Why this step? We start with the first two points; if $A, B, C$ share a line, this slope sets the "target" that the rest must match.
Slope $B C$ : $\frac{11 - 5}{6 - 3} = \frac{6}{3} = 2$ . Why this step? By the similar-triangle logic from the parent note, a line's slope is the same between any pair, so $B C$ must equal $A B$ .
Compare: $2 = 2$ , so the three points are collinear. Why this step? Equal slopes through a shared point ( $B$ ) mean $A$ , $B$ , $C$ cannot bend away from each other — they lie on one line.

Verify: cross-check with $A C$ : $\frac{11 - 1}{6 - 1} = \frac{10}{5} = 2$ . ✓ All three agree — one straight line. (This underpins Equation of a Straight Line.)

Worked example Example 11 — Case L: the indeterminate case (same point twice)

What is the slope "through" $A = (4, 7)$ and $B = (4, 7)$ — the very same point listed twice?

Forecast: in Example 4(E) a zero run gave "undefined". Here both rise and run are zero. Is it the same answer?

Rise $= 7 - 7 = 0$ ; run $= 4 - 4 = 0$ . Why this step? Identical points have no change in either coordinate.
Form the ratio: $m = \frac{0}{0}$ . Why this step? We are honestly plugging into the formula to see what it says.
Read the result: $\frac{0}{0}$ is indeterminate, not merely undefined. Why this step? A single point does not determine a line at all — infinitely many lines pass through one point, each with a different slope. So the formula genuinely cannot pick one answer.

Verify: contrast the two zero-cases. Ex 4(E): $\frac{8}{0}$ (non-zero over zero) = undefined — a real vertical line exists, but its slope has no value. Ex 11: $\frac{0}{0}$ = indeterminate — there is no single line to speak of. Moral: you need two distinct points for slope to make sense. ✓

Recall

Recall Cover and answer

What does the symbol $Δ$ mean? ::: "The change in", measured as final minus initial. Steepness of $m = 2.5$ vs $m = - 2$ — which is steeper? ::: $m = 2.5$ , because $∣2.5∣ > ∣ - 2∣$ . Slope $0$ vs slope undefined — which is horizontal? ::: Slope $0$ is horizontal; undefined is vertical. Line through $(2, 3), (k, 11)$ has slope $4$ — find $k$ . ::: $k = 4$ . Slope of a line with $m = - 1$ — what is its angle of inclination? ::: $135°$ (i.e. $arctan (- 1) = - 45°$ , plus $180°$ ). How do you test if three points lie on one line using slope? ::: Check that every pair of points gives the same slope. Slope of a $45°$ line? ::: $m = tan 45° = 1$ . A ramp rises $0.4$ m over $5$ m — what is its slope? ::: $0.08$ (i.e. $1 : 12.5$ ). Slope "through" one point listed twice? ::: $\frac{0}{0}$ , indeterminate — a single point does not define a line.

Connections

Parent: Slope (gradient) — the definition these examples drill.
Angle of Inclination — Example 7 turns slope into an angle via $arctan$ , including negative and obtuse cases.
Equation of a Straight Line — Example 10's collinearity is one line's equation.
Parallel and Perpendicular Lines — equal slopes (Ex 10) means parallel/collinear.
Rate of Change & Derivative as a Slope — Example 8's ramp gradient generalises to instantaneous slope.
Distance Formula & Midpoint Formula — same $Δ x, Δ y$ machinery.