Use equal distances for bisectors and nearer point half planes
Derive a perpendicular-bisector equation from two complex distances and choose the correct half-plane for a nearer-point condition.
Exam scope and assessment notes
Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027, section 3.9. Paper 1 algebra and radian trigonometry are assumed. Use Cartesian working for products, quotients and the two square roots of a non-zero complex number. Arguments usually use −π < θ ≤ π; an alternative interval 0 ≤ θ < 2π is acceptable unless specified. The argument of zero is undefined. General nth-root methods are beyond this mapped sequence. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Expand squared Cartesian distances and cancel like terms.
- Interpret perpendicular bisectors and linear inequalities.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Why can |z − a| = |z − b| be squared without changing its solutions?
What you will learn
- Derive and sketch a locus equidistant from two distinct points.
- Select the half-plane for a strict or inclusive distance comparison.
Equal distances give a perpendicular bisector
For two distinct fixed points a and b, |z − a| = |z − b| describes their perpendicular bisector. Squaring is valid because both distances are non-negative. With a = −2 + i and b = 4 + i, write (x + 2)² + (y − 1)² = (x − 4)² + (y − 1)². Cancelling gives 12x = 12, so x = 1, the vertical line through their midpoint (1,1). Every point on that line is equidistant, not just the midpoint.
For non-horizontal pairs, expansion still cancels the x² and y² terms and leaves a straight line. With a = 1 + i and b = 3 + 3i, equality gives x + y = 4. The segment between the points has gradient 1, and this bisector has gradient −1 through midpoint (2,2). If a = b, every point is equidistant; the distinct-point condition is needed for a unique bisector.
Choose a side by algebra or a test point
The strict condition |z − a| < |z − b| means z is nearer a. For a = −2 + i and b = 4 + i it gives x < 1. The bisector is excluded; the point a itself satisfies the condition because its distances are zero and six. A test point identifies the side and catches a reversed inequality. Replacing < with ≤ includes the bisector.
Combining a distance comparison with a disk condition means intersecting the two permitted regions. For |z + 2| ≤ |z − 2| and |z| ≤ 3, the first condition is x ≤ 0 and the second is a closed radius-three disk about zero. Their intersection is its left semicircular disk, including the vertical diameter and the circular boundary. Take the intersection of all conditions rather than the union of separately shaded regions.
Pause and explain
For a = −2 + i and b = 4 + i, which region is nearer a?
Worked example
For a = −2 + i and b = 4 + i, find the loci |z − a| = |z − b| and |z − a| < |z − b|.
Show the worked solution
- Square both non-negative distances and cancel the common y and quadratic terms.
- Equality reduces to x = 1, the perpendicular bisector.
- The inequality reduces to x < 1; testing a confirms the nearer-a side and excludes the bisector.
Answer Equality: x = 1. Nearer a: x < 1, boundary excluded.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Derive the locus |z − (1 + i)| = |z − (3 + 3i)| for z = x + iy and check its midpoint and direction.
Give me a hint
Expand the squared distances and cancel x² and y².
Compare my reasoning
- Use (x − 1)² + (y − 1)² = (x − 3)² + (y − 3)².
- Simplification gives 4x + 4y = 16, so x + y = 4.
- It passes through midpoint (2,2) and has gradient −1, perpendicular to the gradient-one joining segment.
x + y = 4; perpendicular bisector through (2,2).
Look for these in your work
- Derive the full line equation from distances.
- Check its midpoint and perpendicular direction.
Derive |z − (−1 + 2i)| = |z − (3 − 2i)| and the strict region nearer 3 − 2i.
Give me a hint
Use the same squared-distance difference, then test the second endpoint.
Compare my reasoning
- Equality is (x + 1)² + (y − 2)² = (x − 3)² + (y + 2)².
- Cancellation gives 8x − 8y = 8, so the bisector is x − y = 1.
- Nearer the second point means its squared distance is smaller, giving x − y > 1; the point (3,−2) confirms this side.
Bisector x − y = 1; nearer 3 − 2i: x − y > 1, boundary excluded.
Look for these in your work
- Use the correct signs for both imaginary coordinates.
- Check the nearer-point side and strict boundary.
Describe the simultaneous conditions |z + 2| ≤ |z − 2| and |z| ≤ 3, including all boundaries.
Give me a hint
Intersect the nearer-point half-plane with the disk rather than combining them by union.
Compare my reasoning
- Squaring the first comparison gives (x + 2)² + y² ≤ (x − 2)² + y², hence x ≤ 0.
- The second condition is x² + y² ≤ 9.
- The intersection is the left semicircular disk of radius 3; both the vertical diameter and its circular arc are included.
x ≤ 0 and x² + y² ≤ 9: left closed semicircular disk, diameter and arc included.
Look for these in your work
- Derive the signed half-plane from the distance comparison.
- Intersect both regions and retain their inclusive boundaries.
Common mistakes
- Equidistance from distinct points defines the entire bisector, not only the midpoint.
- Test a fixed endpoint to check which side is nearer it.
- Combined conditions require intersection and separate boundary checks.
What can you explain now?
If the fixed points a and b coincide, what is |z − a| = |z − b|?
Compare with the explanation
Every point of the plane.
The two distances are identical for every z; there is no unique bisector.
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Make it stick
Derive a bisector from equal complex distances, choose the nearer-point half-plane and intersect it with any further conditions.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.