Hint
The two fixed points are distinct. Average the real components.
Worked solution
(a)
Equate the two squared distances. After cancellation the equation is \(1x+2y=34\), a line through the segment midpoint and perpendicular to the segment.
Answer: \(1x+2y=34\)
(b)
Subtracting the squared distances gives 4(ux+vy−rhs). Being nearer a makes this expression negative. The strict half-plane contains a and excludes the boundary line.
Answer: \(1x+2y<34\)
(c)
Take the negative reciprocal of \(\displaystyle \frac{v}{u}\).
\(\displaystyle - \frac{1}{2}\)
Answer: \(\displaystyle - \frac{1}{2}\)
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: Square and equate the two distances, cancel quadratic terms and simplify to a line.
- 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: The locus is a full bisector, unlike a fixed-argument ray.
(b) · 2 marks
- 1 mark: Compare squared distances and select the strict half-plane containing a.
- 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 strict distance inequality excludes its boundary.
(c) · 2 marks
- 1 mark: Take the negative reciprocal of \(\displaystyle \frac{v}{u}\).
- 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: The locus is a full bisector, unlike a fixed-argument ray.
Total: 6 marks.