Hint
A gradient quotient needs a nonzero horizontal difference. Sum the squared coordinate differences.
Worked solution
(a)
\(5\) follows from Pythagoras; average the coordinates for the midpoint.
Answer: \(5\); \(\displaystyle \left( 13, \ \frac{7}{2}\right)\)
(b)
The gradient is \(\displaystyle \frac{3}{4}\); use the point-gradient form through A.
Answer: \(\displaystyle y - 2 = \frac{3 x}{4} - \frac{33}{4}\)
(c)
Average the y-coordinates.
\(\displaystyle \frac{7}{2}\)
Answer: \(\displaystyle \frac{7}{2}\)
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.