Hint
Use the stated units and precision. Use AP as the stated fraction of the full displacement AB.
Worked solution
(a)
Use AP as the stated fraction of the full displacement AB.
The fraction \(\displaystyle \frac{1}{2}\) of AB adds 6 horizontally to A.
Answer: \(3\)
(b)
Apply the ratio fraction to the vertical displacement.
The same fraction adds 4 vertically to A.
Answer: \(5\)
(c)
The segment vector is a positive scalar multiple of AB and begins at the same point A. Adding it to position(A) gives the stated coordinates of P. The scalar lies strictly between 0 and 1, so P lies inside the segment.
Answer: \(\left[\begin{matrix}3\\5\end{matrix}\right]\); \(\displaystyle AP=\)\(\displaystyle \frac{1}{2}\)AB.
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: Use AP as the stated fraction of the full displacement AB.
- 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: Apply the ratio fraction to the vertical displacement.
- 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) · 3 marks
- 1 mark: Express AP as the correct part-to-whole multiple of AB.
- 1 mark: Add to A and justify collinearity using a shared point and scalar-multiple vectors.
- 1 mark: Give the complete requested result, using the required units, precision and boundary inclusion.
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: 7 marks.