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Year 8 Unit 2 Lesson 9 of 20 +85 XP

Finding Gradient from Two Points

You have two points on a line — you can find the gradient without drawing the graph. The formula $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ turns coordinates into steepness.

Think First: Two points: $(1, 2)$ and $(4, 8)$. How much does $y$ change? How much does $x$ change? What is their ratio?
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Think First · warm-up

Think First
warm-up

You have two points: $(1, 2)$ and $(4, 8)$. How much does $y$ change? How much does $x$ change? What is their ratio?

x y (1, 2) (4, 8) Run = 3 Rise = 6 Ratio = 6/3 = 2

$y$ goes from $2$ to $8$: change in $y = 8 - 2 = 6$. $x$ goes from $1$ to $4$: change in $x = 4 - 1 = 3$. The ratio is $\frac{6}{3} = 2$. This is the gradient of the line!

Learn · 6 cards

1
The Gradient Formula
+5 XP

We already know that gradient $= \dfrac{\text{rise}}{\text{run}}$. Now we turn this into a formula using the coordinates of two points.

The gradient formula is:

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

  • Rise = vertical change = $y_2 - y_1$
  • Run = horizontal change = $x_2 - x_1$
  • Gradient $m = \dfrac{\text{rise}}{\text{run}}$
The Gradient Formula x y A(x1, y1) B(x2, y2) y2-y1 x2-x1
Memory tip: “$y$ on top, $x$ on bottom”
2
Learning Intentions
objectives

Know

  • The gradient formula $m = \dfrac{y_2 - y_1}{x_2 - x_1}$
  • That $(x_1, y_1)$ and $(x_2, y_2)$ are any two points on the line

Understand

  • Why the formula works (rise over run)
  • Why the order of subtraction must be consistent
  • Why a common mistake is reversing the formula

Can Do

  • Calculate gradient given any two points
  • Handle negative coordinates
  • Simplify the result
3
Key Terms
vocabulary
Gradient formula$m = \dfrac{y_2 - y_1}{x_2 - x_1}$, used to calculate gradient through any two known points.
Rise ($\Delta y$)The vertical change: $\Delta y = y_2 - y_1$. Positive = up, negative = down.
Run ($\Delta x$)The horizontal change: $\Delta x = x_2 - x_1$. Always left to right.
Ordered pairA pair $(x, y)$ in specific order. $x$ always comes first.
SubstitutionReplacing variables with known values in the formula.
Consistent orderingUsing the same point as "point 2" in both numerator and denominator.

Watch Me Solve It · 3 examples

Watch Me Solve It · Basic Gradient
+15 XP per step
Q1
PROBLEM
Find the gradient of the line passing through $(2, 3)$ and $(6, 7)$.
x y (2, 3) (6, 7) rise=4 run=4
  1. 1
    Label the points
    Let $(x_1, y_1) = (2, 3)$ and $(x_2, y_2) = (6, 7)$
    It doesn't matter which point is point 1 or 2 — as long as we are consistent.
  2. 2
    Calculate the rise
    Rise $= y_2 - y_1 = 7 - 3 = 4$
  3. 3
    Calculate the run
    Run $= x_2 - x_1 = 6 - 2 = 4$
  4. 4
    Divide rise by run
    $m = \dfrac{4}{4} = 1$
    Gradient = 1. For every 1 unit across, the line goes up 1 unit.
Answer$m = 1$
Watch Me Solve It · Negative Coordinates
+15 XP per step
Q2
PROBLEM
Find the gradient of the line through $(-1, 2)$ and $(3, -4)$.
x y -2 -1 1 2 3 -1 1 2 (-1, 2) (3, -4)
  1. 1
    Label the points
    $(x_1, y_1) = (-1, 2)$ and $(x_2, y_2) = (3, -4)$
  2. 2
    Calculate the rise
    Rise $= (-4) - 2 = -4 - 2 = -6$
  3. 3
    Calculate the run
    Run $= 3 - (-1) = 3 + 1 = 4$
    Subtracting $-1$ means we add $1$.
  4. 4
    Divide rise by run
    $m = \dfrac{-6}{4} = -\dfrac{3}{2} = -1.5$
    Negative gradient — the line goes downhill from left to right.
Answer$m = -1.5$ (or $-\frac{3}{2}$)
4
Consistent Ordering & Why Both Forms Work
+5 XP

Why must $y_2 - y_1$ pair with $x_2 - x_1$? If you mix up the order you get the wrong sign.

Correct
$m = \dfrac{y_2 - y_1}{x_2 - x_1}$ — both in the same direction.
Also correct
$m = \dfrac{y_1 - y_2}{x_1 - x_2}$ — reversed both, so negatives cancel.
Wrong!
$m = \dfrac{y_2 - y_1}{x_1 - x_2}$ — mixed up! Top forward, bottom backward.

This works because $\dfrac{y_1 - y_2}{x_1 - x_2} = \dfrac{-(y_2-y_1)}{-(x_2-x_1)} = \dfrac{y_2-y_1}{x_2-x_1}$. The two negatives cancel.

5
Fractions, Decimals and Simplifying
+5 XP

The gradient formula often gives a fraction. Always simplify your final answer.

As a fraction
$m = \frac{3}{4}$ or write as $0.75$
Simplify!
$\frac{6}{8} = \frac{3}{4}$. Always simplify.
Two negatives
$\frac{-4}{-2} = 2$. Don't leave two negatives!
6
Common Pitfalls
heads-up
Mixing up the order
Using $y_2 - y_1$ in the numerator but $x_1 - x_2$ in the denominator. The subtractions must go in the same direction.
Fix: Label your points before you start, then double-check.
Subtracting a negative incorrectly
Writing $3 - (-1) = 2$ instead of $4$.
Fix: $a - (-b) = a + b$. Use brackets! $3 - (-1) = 3 + 1 = 4$.
Writing the formula upside-down
Using $m = \dfrac{x_2 - x_1}{y_2 - y_1}$ instead of the correct form.
Fix: “$y$ is up high, $x$ is down below.” Rise over run.
Not simplifying the final fraction
Leaving the answer as $\frac{6}{8}$ or $\frac{-4}{-2}$.
Fix: Always simplify! $\frac{6}{8} = \frac{3}{4}$ and $\frac{-4}{-2} = 2$.

Brain Trainer · 10 problems

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Brain Trainer · Gradient Calculations
10 problems

Quick-fire gradient calculations. Set a timer for 3 minutes!

  1. 1 Find the gradient between $(1, 1)$ and $(5, 9)$.

    $m = \frac{9-1}{5-1} = \frac{8}{4} = 2$m = 2

  2. 2 Find the gradient between $(0, 0)$ and $(3, -6)$.

    $m = \frac{-6-0}{3-0} = \frac{-6}{3} = -2$m = -2

  3. 3 Find the gradient between $(-2, 4)$ and $(2, 4)$.

    $m = \frac{4-4}{2-(-2)} = \frac{0}{4} = 0$ (horizontal line)m = 0

  4. 4 Find the gradient between $(3, -1)$ and $(7, -5)$.

    $m = \frac{-5-(-1)}{7-3} = \frac{-4}{4} = -1$m = -1

  5. 5 Find the gradient between $(-4, -3)$ and $(-1, 6)$.

    $m = \frac{6-(-3)}{-1-(-4)} = \frac{9}{3} = 3$m = 3

  6. 6 Find the gradient between $(5, 2)$ and $(5, 8)$.

    $m = \frac{8-2}{5-5} = \frac{6}{0} =$ undefined (vertical line)Undefined

  7. 7 Find the gradient between $(-3, 0)$ and $(6, -6)$. Simplify fully.

    $m = \frac{-6-0}{6-(-3)} = \frac{-6}{9} = -\frac{2}{3}$m = -2/3

  8. 8 True or False: Swapping which point you call $(x_1, y_1)$ always changes the gradient.

    False. As long as you are consistent, swapping labels gives the same answer. $\frac{-a}{-b} = \frac{a}{b}$.

  9. 9 Find the gradient between $(-5, -2)$ and $(-1, -8)$.

    $m = \frac{-8-(-2)}{-1-(-5)} = \frac{-6}{4} = -\frac{3}{2}$m = -3/2

  10. 10 A line has gradient $4$ and passes through $(2, 1)$. Use the gradient to find another point on the line.

    Gradient 4 means "rise 4, run 1". From $(2, 1)$: go right 1 and up 4.(3, 5)

Quick Check · 5 questions

1
What is the gradient between $(1, 3)$ and $(4, 9)$?
+10 XP
2
Find the gradient between $(0, 0)$ and $(-2, 6)$.
+10 XP
3
Two points on a line are $(3, 5)$ and $(7, 5)$. What is the gradient?
+10 XP
4
What is a common mistake with the gradient formula?
+10 XP
5
Find the gradient between $(-3, -2)$ and $(1, 4)$.
+10 XP

Show Your Working · 3 questions

Show Your Working
written questions
Apply Medium

SAQ 1. Find the gradient of the line passing through $(2, 5)$ and $(8, 11)$. Show all working.

Answer in your workbook.
Understand Medium

SAQ 2. Explain why $m = \frac{y_1-y_2}{x_1-x_2}$ gives the same answer as $m = \frac{y_2-y_1}{x_2-x_1}$. Use an example.

Answer in your workbook.
Apply Hard

SAQ 3. A line has gradient $3$ and passes through $(1, 2)$. Find another point on the line and explain your reasoning.

Answer in your workbook.
Comprehensive Answers

Quick Check

1. C — $m = 2$.

2. C — $m = -3$.

3. C — $m = 0$ (horizontal line).

4. B — Upside-down formula.

5. D — $m = \frac{3}{2}$.

Written Questions Model Answers

SAQ 1: Let $(x_1,y_1) = (2,5)$ and $(x_2,y_2) = (8,11)$. $m = \frac{11-5}{8-2} = \frac{6}{6} = 1$.

SAQ 2: Both forms give the same answer because when you reverse the subtraction in both numerator and denominator, you multiply both by $-1$. $\frac{-a}{-b} = \frac{a}{b}$. Example using $(2,3)$ and $(6,7)$: $\frac{7-3}{6-2} = \frac{4}{4} = 1$ and $\frac{3-7}{2-6} = \frac{-4}{-4} = 1$. Both give $m=1$.

SAQ 3: Gradient $3$ means "rise 3, run 1". From $(1,2)$: go right $1$ to $x=2$, go up $3$ to $y=5$. Another point is $(2,5)$. Check: $m = \frac{5-2}{2-1} = 3$. Other valid answers: $(0,-1)$, $(3,8)$, etc.

Stretch Challenge · +25 XP, +10 coins

Collinear Points and Reverse Gradient

Part A: Find the gradient of the line through each pair of points: (a) $(0,-3)$ and $(4,5)$   (b) $(-2,-5)$ and $(-2,3)$   (c) $(-4,7)$ and $(2,-2)$

Part B: Three points are $A(1,2)$, $B(3,8)$, $C(5,14)$. Find $m_{AB}$, $m_{BC}$ and $m_{AC}$. What do you notice? What does this tell you about the points?

Part C: A line has gradient $-\frac{2}{3}$ and passes through $(6,5)$. By going backwards (reverse the gradient), find the point where this line crosses the $y$-axis.

Reveal solution

Part A: (a) $m = \frac{5-(-3)}{4-0} = 2$. (b) $m = \frac{3-(-5)}{-2-(-2)} = \frac{8}{0}$ = undefined (vertical line). (c) $m = \frac{-2-7}{2-(-4)} = -\frac{3}{2}$.

Part B: $m_{AB} = m_{BC} = m_{AC} = 3$. All gradients are equal — the three points are collinear (on the same straight line).

Part C: Gradient $-\frac{2}{3}$ means for every 3 left, go up 2. From $x=6$ to $x=0$ is 6 left = two steps, so up $4$. $y = 5 + 4 = 9$. The line crosses the $y$-axis at $(0, 9)$.

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Key Takeaways
  • The gradient formula is $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ for any two points $(x_1,y_1)$ and $(x_2,y_2)$.
  • Label your points first, then substitute carefully — keep the same ordering in numerator and denominator.
  • Use brackets when substituting negative coordinates: $3 - (-1) = 3 + 1 = 4$.
  • Positive gradient means the line slopes uphill; negative means downhill.
  • A gradient of zero means a horizontal line; undefined gradient means a vertical line.
  • Always simplify the final fraction.
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Lesson Complete!

You've mastered finding gradient from two points.

+50 XP