Hint
Use the stated units and precision. Apply Pythagoras to the coordinate differences.
Worked solution
(a)
Apply Pythagoras to the coordinate differences.
The squared distance is 100, so \(\displaystyle AB=10\).
Answer: \(10\)
(b)
Average the two x-coordinates.
\(3\) is the mean of the endpoint x-coordinates.
Answer: \(3\)
(c)
Average the two y-coordinates.
\(2\) is the mean of the endpoint y-coordinates.
Answer: \(2\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternative methods and consequential reasoning. An unexplained result does not earn a method mark.
(a) · 2 marks
- 1 mark: Apply Pythagoras to the coordinate differences.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Keep intermediate values exact until the final step.
(b) · 2 marks
- 1 mark: Average the two x-coordinates.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Check the answer against the original conditions.
(c) · 2 marks
- 1 mark: Average the two y-coordinates.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Keep intermediate values exact until the final step.
Total: 6 marks.