Distance, midpoint & gradient
AQA GCSE Maths revision on Distance, midpoint & gradient. Aligned to the AQA GCSE Mathematics 8300 specification. This bank has 19 practice questions on this topic.
Sample questions (3 of 19)
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Question 1
A student calculates the gradient of a line by doing (x2 - x1) / (y2 - y1). What is the fundamental error in this approach?
- A) The coordinates are subtracted in the wrong order.
- B) The ratio is inverted compared to the definition of rise over run.
- C) The formula requires addition instead of subtraction.
- D) The student has ignored the negative signs of the coordinates.
Show answer
Answer: The ratio is inverted compared to the definition of rise over run.
The gradient represents the steepness of a line, calculated as the vertical change (rise) divided by the horizontal change (run). By calculating (x2 - x1) / (y2 - y1), the student has calculated the reciprocal of the gradient, which represents the run over the rise instead.
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Question 2
When calculating the distance between two points (x1, y1) and (x2, y2), which geometric principle is being applied?
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Answer: Pythagoras' theorem
The horizontal distance and vertical distance between two points form the two shorter sides of a right-angled triangle. The distance between the points is the hypotenuse, which is found using Pythagoras' theorem: a squared plus b squared equals c squared.
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Question 3
Which of these lines has a gradient that is undefined?
- A) A horizontal line.
- B) A vertical line.
- C) A line passing through the origin.
- D) A line with a gradient of zero.
Show answer
Answer: A vertical line.
Gradient is calculated as rise divided by run. For a vertical line, the 'run' (change in x) is zero, and division by zero is mathematically undefined in the context of gradients.
Want to test yourself on the remaining cards for this topic?
16 more questions on Distance, midpoint & gradient — plus mistakes tracking and spaced repetition across the whole Maths spec.