Reflections in horizontal and vertical lines
Reflect points at equal perpendicular distances from a stated mirror line.
Section Paper 1 · Core · Non-calculator
Core and Extended. Core reflections use horizontal or vertical lines; other straight mirror lines are Extended. This is the separate 9–1 qualification 0980. Cambridge intends Core for grades 4–1 and Extended for grades 9–5; the eligible award bands are grades 5–1 for the Core papers and grades 9–3 for the Extended papers.
Before you begin
- Plot coordinates.
- Measure perpendicular distances on a grid.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What equation names the x-axis?
What you will learn
- Reflect a polygon in x = a or y = b.
- Describe a reflection by naming its mirror line.
Equal distances on opposite sides
A reflected point and its image lie the same perpendicular distance from the mirror line, on opposite sides. A point on the mirror line stays fixed. Reflection preserves lengths and angles but reverses orientation; the mirror line is the perpendicular bisector of each point–image segment.
In x = 2, point (5,1) is three right of the mirror and maps to (−1,1), three left. The horizontal coordinate changes while the vertical coordinate stays the same. Join reflected polygon vertices in their original order rather than changing their names.
Read the line equation before changing coordinates
For reflection in x = a, the image is (2a − x, y). For reflection in y = b, it is (x, 2b − y). Thus reflection of (3,−2) in y = 1 gives (3,4). The x-axis is y = 0, while the y-axis is x = 0; confusing these names reflects in the wrong direction.
To identify the mirror from a pair of corresponding points, find their midpoint and the perpendicular bisector. With horizontal movement only, the mirror is vertical through the midpoint. A full description says reflection in a named line such as x = 2, not merely 'flipped left'.
Pause and explain
Reflect (3,−2) in y = 1.
Worked example
Reflect (5,1) in x = 2.
Show the worked solution
- The original is three units to the right of x = 2.
- Move three units left of the mirror to x = −1.
- The vertical coordinate stays one; both points have equal mirror distance.
Answer (−1,1).
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 (4,−3) in the y-axis.
Give me a hint
The y-axis is the vertical line x = 0.
Compare my reasoning
- The point is four units right of the mirror.
- The image is four left with the same y-coordinate.
- Its coordinates are (−4,−3).
(−4,−3).
Look for these in your work
- Identified the correct axis.
- Preserved equal perpendicular distance.
Reflect A(−1,2), B(2,2), C(0,4) in y = 3.
Give me a hint
Use y′ = 6 − y for each vertex.
Compare my reasoning
- A′ = (−1,4) and B′ = (2,4).
- C′ = (0,2).
- Join the images in order; each pair has midpoint on y = 3.
A′(−1,4), B′(2,4), C′(0,2).
Look for these in your work
- Applied the same mirror to all vertices.
- Checked pair midpoints.
Point (−2,5) reflects to (6,5). Find the mirror line.
Give me a hint
The joining segment is horizontal.
Compare my reasoning
- Its midpoint is (2,5).
- The perpendicular bisector is vertical.
- The mirror is x = 2, four units from each point.
Reflection in x = 2.
Look for these in your work
- Used the midpoint.
- Gave the complete line equation.
Common mistakes
- Confusing x = a with a horizontal line.
- Measuring a sloping distance to the mirror.
- Giving an incomplete reflection description.
What can you explain now?
What happens to a point on the mirror line?
Compare with the explanation
It stays fixed.
Its perpendicular mirror distance is zero, so its reflected position is unchanged.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Tomorrow, reflect one triangle in each type of line and verify equal perpendicular distances.
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.