Hint
The height OH is not vertical. Find its line using a perpendicular gradient, then solve for its intersection with AB.
Worked solution
(a)
The intercepts give x/6+y/8=1, hence 4x+3y=24. The endpoint differences are −6 and 8, so AB=√(36+64)=10.
Answer: \(4x+3y=24\); \(AB=10\).
(b)
AB has gradient −4/3, so OH has gradient 3/4 and equation y=3x/4. Substitution gives 4x+9x/4=24, so x=96/25 and y=72/25.
Answer: \(H=(96/25,72/25)\).
(c)
OH=√(96²+72²)/25=120/25=24/5. Using AB as the base, the area is (1/2)(10)(24/5)=24 square coordinate units.
Answer: \(OH=24/5\); area \(24\).
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 3 marks
- 1 mark: Use the points or intercepts to obtain a valid line method.
- 1 mark: Give 4x+3y=24.
- 1 mark: Obtain AB=10.
Alternative methods: A point-gradient line calculation is accepted.
Common error: Taking an axis intercept as the sloping side length.
(b) · 3 marks
- 1 mark: Use the perpendicular line y=3x/4 through O.
- 1 mark: Solve its intersection with AB correctly.
- 1 mark: Give both coordinates (96/25,72/25).
Alternative methods: An orthogonal projection or equivalent simultaneous-equation method is accepted.
Common error: Using a median from O instead of the perpendicular height.
(c) · 3 marks
- 1 mark: Obtain exact height OH=24/5.
- 1 mark: Use one half times AB times OH.
- 1 mark: Obtain area 24 square coordinate units; method credit may follow through a previously calculated positive base/height.
Alternative methods: The requested height/base calculation is needed; the axis triangle area can check the result.
Common error: Using the full distance from O to an endpoint as the height.
Total: 9 marks.