Hint
A gradient quotient needs a nonzero horizontal difference. Sum the squared coordinate differences.
Worked solution
(a)
\(\sqrt{41}\) follows from Pythagoras; average the coordinates for the midpoint.
Answer: \(\sqrt{41}\); \(\displaystyle \left( \frac{29}{2}, \ 5\right)\)
(b)
The gradient is \(\displaystyle \frac{4}{5}\); use the point-gradient form through A.
Answer: \(\displaystyle y - 3 = \frac{4 x}{5} - \frac{48}{5}\)
(c)
Average the y-coordinates.
\(5\)
Answer: \(5\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternatives and consequential reasoning; an unexplained final answer does not earn a method mark.
(a) · 2 marks
- 1 mark: Use coordinate differences for distance and coordinate averages for the midpoint.
- 1 mark: Give all results requested in this part, with the correct signs, exact values, units, branch and boundary inclusion where applicable.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: A midpoint is an average, not a difference.
(b) · 2 marks
- 1 mark: Use the two-point gradient and one of the stated points.
- 1 mark: Give all results requested in this part, with the correct signs, exact values, units, branch and boundary inclusion where applicable.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Do not interchange the horizontal and vertical differences in a gradient.
(c) · 2 marks
- 1 mark: Average the 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: A midpoint is an average, not a difference.
Total: 6 marks.