Reflect in a mirror line using equal perpendicular distances
Locate a reflected point across a specified line · Use vertical and horizontal line rules · Recognise points fixed on the mirror
Exam scope and assessment notes
Core and Extended foundations. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.
Before you begin
- Perpendiculars and midpoints
- Coordinates
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should you establish before applying the method in this lesson?
What you will learn
- Locate a reflected point across a specified line
- Use vertical and horizontal line rules
- Recognise points fixed on the mirror
The mirror bisects the joining segment
For reflection in a line, the original and image lie at equal perpendicular distances on opposite sides. If F is the perpendicular foot from P onto the mirror, P′ = 2F−P in coordinates. A point on the mirror remains fixed. Reflection preserves lengths and angles but reverses orientation.
Axis-aligned rules
Reflection in x=a sends (x,y) to (2a−x,y); reflection in y=b sends it to (x,2b−y). The x-axis is y=0, so it changes the y sign. The y-axis is x=0, so it changes the x sign.
Describe a reflection by its mirror line. A line joining one original point to its image is generally perpendicular to that mirror, not the mirror itself.
Sources for this lesson
Pause and explain
Which coordinate changes sign in reflection in the x-axis?
Worked example
Reflect A(2,1), B(4,1), C(2,3) in x=1.
Show the worked solution
- Use x′ = 2−x and keep y unchanged.
- Images are A′(0,1), B′(−2,1), C′(0,3).
Answer A′(0,1), B′(−2,1), C′(0,3).
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.
Reflect (3,−2) in x=1.
Give me a hint
Use twice the mirror coordinate minus x.
Compare my reasoning
- x′ = 2−3 = −1.
- y stays −2.
(−1,−2).
Look for these in your work
- Use the stated mirror
- Keep the other coordinate
P(4,−1) maps to P′(4,5) by reflection in a horizontal line. Find the mirror.
Give me a hint
The mirror is halfway between the y coordinates.
Compare my reasoning
- Midpoint y = (−1+5)/2 = 2.
- A horizontal line has equation y=2.
Mirror y=2.
Look for these in your work
- Use perpendicular midpoint geometry
- Name the full line equation
The perpendicular foot from P(4,6) to a mirror is F(1,2). Find its image and verify F is the midpoint.
Give me a hint
Use P′ = 2F−P.
Compare my reasoning
- P′ = (2−4,4−6) = (−2,−2).
- Midpoint of P and P′ is (1,2), the given foot.
P′(−2,−2).
Look for these in your work
- Use the supplied perpendicular foot
- Check equal opposite distances
Common mistakes
- The x-axis changes y, not x.
- Equal distance must be perpendicular to the mirror.
What can you explain now?
Which coordinate changes sign in reflection in the x-axis?
Compare with the explanation
The y coordinate.
The mirror is y=0, so x stays unchanged.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Which coordinate changes sign in reflection in the x-axis?
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.