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For z=x+yi, derive the Cartesian equation of its locus.

Read the idea, work independently, then explain what changed.

TOPIC 01

2022 JM02

This is the perpendicular bisector of the segment joining z₁ and z₂.

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Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Standard#Your turn

For z=x+yi, derive the Cartesian equation of its locus.

z1=3+5i,z2=5+i,∣z−z1∣=∣z−z2∣z_1=3+5i,\quad z_2=5+i,\quad|z-z_1|=|z-z_2|

Official paper · jm02-2022 · 4(a)(i) · PDF 6

Official original and suggested answers ↗ · Suggested answer PDF page 11

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Write the two distances as square roots.
Hint 2
Square and cancel x² and y².
Worked solution
  1. Equal distances give equal squared distances.

    (x−3)2+(y−5)2=(x−5)2+(y−1)2(x-3)^2+(y-5)^2=(x-5)^2+(y-1)^2
  2. Collect the linear terms.

    4x−8y+8=0  ⟺  x−2y+2=04x-8y+8=0\iff x-2y+2=0

x−2y+2=0.

Checks and common pitfalls: This is the perpendicular bisector of the segment joining z₁ and z₂.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Equal distances give equal squared distances.
    (x−3)2+(y−5)2=(x−5)2+(y−1)2(x-3)^2+(y-5)^2=(x-5)^2+(y-1)^2
  • Collect the linear terms.
    4x−8y+8=0  ⟺  x−2y+2=04x-8y+8=0\iff x-2y+2=0

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Curriculum and source notes ↗