A race timer records a horse's position at two checkpoints. A price list shows the cost of hay at two different quantities. Two data points are all you need — the slope formula extracts the rate of change from any pair of coordinates.
The slope formula:
$$m = \frac{y_2 - y_1}{x_2 - x_1}$$
Label one point $(x_1, y_1)$ and the other $(x_2, y_2)$. The order does not matter — as long as you are consistent (subtract in the same order top and bottom). A negative result means the line falls left-to-right. A fractional result is fine — leave it as a fraction or convert to a decimal.
Worked Examples
Worked Example 1 — Positive slope
Find the slope through $(2, 5)$ and $(6, 13)$.
$$m = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2$$
Worked Example 2 — Negative slope
Find the slope through $(1, 9)$ and $(4, 3)$.
$$m = \frac{3 - 9}{4 - 1} = \frac{-6}{3} = -2$$
Worked Example 3 — Find a missing coordinate
The slope through $(2, y)$ and $(5, 11)$ is $3$. Find $y$.
$$m = \frac{11 - y}{5 - 2} = 3$$
$$\frac{11 - y}{3} = 3 \quad | \times 3$$
$$11 - y = 9 \quad | -11$$
$$-y = -2 \quad | \times (-1)$$
$$y = 2$$
Warm-Up — Compute slope from two points (formula shown)
-
Problem 1
Find the slope through $(0, 1)$ and $(4, 9)$.
$m = \dfrac{9 - 1}{4 - 0}$
calculate →
-
Problem 2
Find the slope through $(3, 10)$ and $(7, 2)$.
$m = \dfrac{2 - 10}{7 - 3}$
calculate → is it positive or negative? →
-
Problem 3
A horse runs a race. At the $200$ m checkpoint it has been running for $20$ seconds. At the $500$ m checkpoint, $50$ seconds have passed. The points are $(20, 200)$ and $(50, 500)$ where $x =$ time (s) and $y =$ distance (m). Find the slope. What does it represent?
$m = \dfrac{500 - 200}{50 - 20}$
calculate → interpret units →
Core Problems — find slope and missing coordinates
-
Problem 4
Find the slope through each pair of points. State whether each line rises or falls.
(a) $(−1, 4)$ and $(3, 12)$
(b) $(2, 7)$ and $(5, 1)$
(c) $(0, 0)$ and $(6, −9)$
apply formula to each → state direction →
-
Problem 5
The slope through $(x, 3)$ and $(9, 15)$ is $4$. Find $x$.
write slope formula → rearrange → solve →
-
Problem 6
The slope through $(2, y)$ and $(8, 5)$ is $-\frac{1}{2}$. Find $y$.
write slope formula → rearrange → solve →
From here — write the equation yourself, then solve
-
Problem 7
A bale of hay costs €18 for 3 kg and €42 for 7 kg. Model this as two points $(x, y)$ where $x$ is kilograms and $y$ is price in euros. Find the slope. What does it tell you about the price?
write points → apply formula → interpret →
-
Problem 8
A riding instructor charges for lessons. At $2$ lessons per week, a student pays €60/week. At $5$ lessons per week, they pay €135/week. Model as two points, find slope, and interpret: what is the cost per extra lesson?
define points → find slope → interpret as €/lesson →
-
Problem 9 Challenge
Three points are given: $A(1, 3)$, $B(4, 9)$, $C(7, 15)$.
(a) Find the slope from $A$ to $B$.
(b) Find the slope from $B$ to $C$.
(c) Are the three points on the same line? How do you know?
(d) If so, find the $y$-value of the point on this line at $x = 10$.
two slopes → compare → extend →
-
Problem 10 Open
Write your own real-world two-point scenario where the slope equals $-2$. Your scenario should have a sensible context (not just "here are two points"). Describe what the negative slope means in your context.
choose context → pick two points with slope −2 → explain sign →
Show answers
Problem 2
$m = -8/4 = -2$ · Negative — line falls from left to right.
Problem 3
$m = 300/30 = 10$ · The slope represents the horse's speed: $10$ metres per second.
Problem 4
(a) $m = (12-4)/(3-(-1)) = 8/4 = 2$ — rises · (b) $m = (1-7)/(5-2) = -6/3 = -2$ — falls · (c) $m = (-9-0)/(6-0) = -9/6 = -3/2$ — falls
Problem 5
$\frac{15-3}{9-x} = 4$ → $12 = 4(9-x)$ → $12 = 36 - 4x$ → $4x = 24$ → $x = 6$
Problem 6
$\frac{5-y}{8-2} = -\frac{1}{2}$ → $\frac{5-y}{6} = -\frac{1}{2}$ → $5-y = -3$ → $y = 8$
Problem 7
Points: $(3, 18)$ and $(7, 42)$ · $m = (42-18)/(7-3) = 24/4 = 6$ · The hay costs €6 per kilogram.
Problem 8
Points: $(2, 60)$ and $(5, 135)$ · $m = (135-60)/(5-2) = 75/3 = 25$ · Each additional lesson costs €25.
Problem 9
(a) $m_{AB} = (9-3)/(4-1) = 6/3 = 2$ · (b) $m_{BC} = (15-9)/(7-4) = 6/3 = 2$ · (c) Yes — same slope means collinear. (d) From $C(7,15)$: $y = 15 + 2(10-7) = 15+6 = 21$
Problem 10
Open — example: "A water trough drains. After 1 minute it has 20 litres, after 3 minutes it has 16 litres." Points $(1,20)$ and $(3,16)$: $m = -4/2 = -2$. Negative means the volume is decreasing — 2 litres lost per minute.
Coming up next → Lesson 18: Rate of Change from Tables and Graphs
Slope is rate of change. In Lesson 18, you'll read rate of change from data tables and graphs — and learn to tell at a glance whether a relationship is linear (constant rate) or not.