Practice Maths

Gradient and Slope

Key Ideas

Key terms: gradient, slope, rise, run, positive gradient, negative gradient, zero gradient, undefined gradient

• Gradient measures the steepness of a line: gradient = rise ÷ run.
• Rise = vertical change (up is positive, down is negative).
• Run = horizontal change (right is positive, left is negative).
• Positive gradient: line goes up from left to right (rise and run have the same sign).
• Negative gradient: line goes down from left to right (rise and run have opposite signs).
• Zero gradient: horizontal line (rise = 0).
• Undefined gradient: vertical line (run = 0, division by zero is undefined).
• From two points: gradient m = (y2 − y1) ÷ (x2 − x1).

Gradient from two points:

The formula is: m = Δy ÷ Δx = (y2 − y1) ÷ (x2 − x1)

Hot Tip Always be consistent with which point you call “Point 1” and “Point 2”. If you subtract y2 − y1, then you must also subtract x2 − x1 (in the same order). The answer will be the same either way, as long as you are consistent.

Worked Example

Question: Find the gradient of the line passing through (1, 2) and (4, 8).

Step 1 — Label the points.
Let (x1, y1) = (1, 2) and (x2, y2) = (4, 8).

Step 2 — Calculate rise and run.
Rise = y2 − y1 = 8 − 2 = 6
Run = x2 − x1 = 4 − 1 = 3

Step 3 — Divide.
m = rise ÷ run = 6 ÷ 3 = 2

Step 4 — Interpret.
The gradient is 2 (positive), so the line rises 2 units for every 1 unit it moves to the right.

What Is Gradient?

The gradient (also called slope) measures the steepness of a line. It tells you how much y changes for every 1 unit increase in x.

Gradient = rise ÷ run

where rise is the vertical change and run is the horizontal change between two points on the line. Pick any two points — the gradient will be the same no matter which two you choose (that's what makes it a straight line).

Example: A line passes through (2, 3) and (6, 11).
Rise = 11 − 3 = 8. Run = 6 − 2 = 4. Gradient = 8 ÷ 4 = 2.

Positive, Negative, Zero, and Undefined Gradient

Positive gradient: The line goes up from left to right (like walking uphill). y increases as x increases. e.g. gradient = 3.

Negative gradient: The line goes down from left to right (like walking downhill). y decreases as x increases. e.g. gradient = −2.

Zero gradient: The line is perfectly horizontal (flat). y stays the same as x changes. The rule looks like y = 5 (no x term). Gradient = 0/run = 0.

Undefined gradient: The line is perfectly vertical. x stays the same as y changes. This is like dividing by zero — gradient = rise/0 which is undefined. Vertical lines are written as x = 3 (no y term).

Gradient in Real Life

Gradient is used to describe the steepness of real-world slopes:

• Roads: A road with gradient 1/10 rises 1 m for every 10 m of horizontal distance. This is an 8.1° slope — quite steep for a road.
• Roofs: A roof pitch is often described as a ratio of rise to run. A 1:3 pitch means for every 3 m across, the roof rises 1 m.
• Wheelchair ramps: Australian standards require a maximum gradient of 1:14 for accessibility — very gentle.
• Ski runs: A black diamond run might have a gradient of 1:2 or steeper.

Calculating Gradient from Two Points

Given two points (x1, y1) and (x2, y2):
m = (y2 − y1) ÷ (x2 − x1)

Example: Points (−1, 4) and (3, −4).
m = (−4 − 4) ÷ (3 − (−1)) = −8 ÷ 4 = −2. Negative — line goes downhill left to right.

Note: you can subtract in either order, as long as you are consistent (same point first in numerator and denominator).

Steeper Lines Have Larger Gradient Values

A gradient of 5 is steeper than a gradient of 2, which is steeper than a gradient of 0.5. For negative gradients, −4 is steeper (falling faster) than −1.

On a graph, a line with gradient 4 rises 4 units for every 1 unit across. A line with gradient 1/2 only rises 1 unit for every 2 units across — much gentler.

Key tip: When calculating gradient, always go from left to right. Count the run in the positive x direction, and count the rise going up as positive and down as negative. This consistent approach prevents sign errors, especially for negative gradients.

Mastery Practice

  1. Calculate the gradient of the line passing through each pair of points. Fluency

    1. (0, 0) and (3, 6)
    2. (1, 3) and (4, 9)
    3. (2, 5) and (6, 13)
    4. (0, 4) and (5, −1)
    5. (1, 7) and (3, 3)
    6. (−2, 1) and (2, 9)
    7. (3, 3) and (7, 3)
    8. (−1, −3) and (2, 6)
    9. (0, 10) and (4, 2)
    10. (−3, 5) and (1, −3)

  2. For each gradient value, classify it as positive, negative, zero, or undefined. Fluency

    1. m = 3
    2. m = −5
    3. m = 0
    4. m = ½
    5. m = −¾
    6. A vertical line
    7. m = −100
    8. m = 0.01

  3. For each description, state whether the gradient is positive, negative, zero, or undefined, then estimate or calculate the value where possible. Fluency

    1. A road that climbs 8 m for every 100 m travelled horizontally.
    2. A horizontal footpath.
    3. A ski slope that drops 15 m for every 30 m of horizontal distance.
    4. A flagpole standing straight up.
    5. A ramp that rises 1 m for every 4 m of horizontal distance.
    6. A line going down from left to right.

  4. Compare the steepness of lines with the given gradients. For each pair, state which line is steeper and explain why. Understanding

    1. m = 3 and m = 5
    2. m = −4 and m = 2
    3. m = ½ and m = ¼
    4. m = −6 and m = −2
    5. m = 1 and m = −1
    6. m = 0.75 and m = ¾

  5. Use the given gradient and one point to find the missing coordinate. Understanding

    1. m = 2; points (1, 3) and (4, ?)
    2. m = −3; points (0, 5) and (2, ?)
    3. m = 4; points (?, 6) and (3, 14)
    4. m = ½; points (2, 1) and (6, ?)
    5. m = −2; points (1, 8) and (?, 2)
    6. m = 3; points (0, ?) and (4, 13)

  6. Gradient in real-world contexts. Problem Solving

    1. A wheelchair ramp must have a maximum gradient of 1:12 (rise of 1 for every run of 12). A ramp needs to rise 0.6 m. What is the minimum horizontal length the ramp must be?
    2. Three points are given: A(0, 2), B(3, 8), and C(6, 14). Show that A, B, and C are collinear (lie on the same straight line) by comparing the gradients of AB and BC.
    3. A road rises from an elevation of 45 m at point A to 165 m at point B. The horizontal distance between A and B is 600 m. Find the gradient of the road. Express it as a simplified fraction and as a decimal.
    4. Line 1 passes through (0, 3) and (4, 11). Line 2 passes through (−2, 9) and (2, 1). Without drawing the lines, describe how they compare in terms of gradient. What do you notice?
  7. Ramp design. A builder is designing two ramps to access a loading dock 1.2 m above ground level. Ramp A has a horizontal run of 6 m. Ramp B has a gradient of 1:8 (rise:run).
    1. Calculate the gradient of Ramp A as a simplified fraction and as a decimal.
    2. What is the horizontal run of Ramp B?
    3. Building regulations state the maximum gradient for a ramp is 1:5. Which ramp(s) comply? Justify your answer.
    Problem Solving

  8. The graph below shows the line y = 2x − 1 with two points marked. Use the marked points to calculate the gradient. Understanding

    -5 -4 -3 -2 -1 1 2 3 4 5 5 4 3 2 1 -1 -2 -3 -4 -5 x y y = 2x − 1 (1, 1) (3, 5)
    1. Write the coordinates of the two marked points.
    2. Calculate the rise (vertical change) and the run (horizontal change) between the two points.
    3. Calculate the gradient using rise ÷ run.
    4. Choose a different pair of points on the line and verify that you get the same gradient.

  9. The graph below shows two lines with different gradients. Study the graph and answer the questions. Problem Solving

    -5 -4 -3 -2 -1 1 2 3 4 5 5 4 3 2 1 -1 -2 -3 -4 -5 x y A: y = 3x B: y = x/2
    1. Which line has the greater gradient? How can you tell just by looking at the graph?
    2. Calculate the gradient of Line A (y = 3x) using two points you can read from the graph. Show your working.
    3. Calculate the gradient of Line B (y = x/2) using two points you can read from the graph. Show your working.
    4. Explain in your own words why a larger gradient value produces a steeper line.

  10. The graph below shows a line with a negative gradient. Use the marked points to answer the questions. Problem Solving

    -5 -4 -3 -2 -1 1 2 3 4 5 5 4 3 2 1 -1 -2 -3 -4 -5 x y y = −2x + 1 (0, 1) (2, −3)
    1. Use the two marked points (0, 1) and (2, −3) to calculate the gradient. Show your working using the formula m = (y2 − y1) ÷ (x2 − x1).
    2. What does a negative gradient tell you about the direction of the line? Describe what you see on the graph.
    3. A different line has gradient −3. Is it steeper or less steep than this line? Explain your reasoning.
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