Reflections in diagonal mirror lines
Reflect in straight lines beyond the Core horizontal and vertical cases.
Extended only. 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
- Reflect in horizontal and vertical lines.
- Plot diagonal lines on a grid.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What happens to a point on a diagonal mirror?
What you will learn
- Reflect coordinates in y = x and y = −x.
- Use perpendicular distances for a general straight mirror line.
Coordinate swaps describe two familiar diagonals
Reflection in y = x maps (x,y) to (y,x), so (2,5) becomes (5,2). Reflection in y = −x maps it to (−y,−x), so (2,5) becomes (−5,−2). Points already on the chosen mirror remain unchanged.
Check that the point–image segment is perpendicular to the mirror and that its midpoint lies on the mirror. Swapping coordinates alone only works for y = x; a translated or differently sloping mirror requires its actual perpendicular geometry.
Construct a reflection without guessing the shift
For another straight line, draw a perpendicular from each vertex to the mirror. Continue the same distance beyond the line to mark its image. Join images with straight edges and verify that the lengths and angle measures are preserved.
A grid can make visual guessing tempting, especially on a line such as y = x + 1. Point (3,1) is three units below that line at x = 3, but the relevant distance is perpendicular, not vertical. Its reflection is (0,4), whose midpoint (1.5,2.5) lies on y = x + 1.
Pause and explain
Reflect (2,5) in y = x.
Worked example
Reflect (2,5) in y = −x.
Show the worked solution
- Use (x,y) → (−y,−x).
- The image is (−5,−2).
- The midpoint (−1.5,1.5) lies on y = −x and the joining segment is perpendicular.
Answer (−5,−2).
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 (−3,4) in y = x.
Give me a hint
Swap the coordinates.
Compare my reasoning
- The original coordinates are x = −3 and y = 4.
- Swap to obtain (4,−3).
- Their midpoint lies on y = x.
(4,−3).
Look for these in your work
- Used the correct diagonal rule.
- Checked the midpoint.
Reflect (−3,4) in y = −x.
Give me a hint
Swap and negate both components.
Compare my reasoning
- The new x is −4.
- The new y is 3.
- The image is (−4,3), with midpoint on y = −x.
(−4,3).
Look for these in your work
- Negated the swapped components.
- Verified the correct mirror.
Show that (3,1) and (0,4) are a reflected pair in y = x + 1.
Give me a hint
Check the midpoint and joining-line direction.
Compare my reasoning
- The midpoint is (1.5,2.5), satisfying y = x + 1.
- The joining segment has gradient −1, perpendicular to mirror gradient one.
- The mirror therefore bisects the segment perpendicularly.
They are a reflected pair in y = x + 1.
Look for these in your work
- Checked both reflection conditions.
- Used perpendicular rather than vertical distance.
Common mistakes
- Using the y = x swap on every diagonal.
- Measuring vertical instead of perpendicular distance.
- Forgetting fixed points on the mirror.
What can you explain now?
Is equal vertical distance enough for a diagonal reflection?
Compare with the explanation
No.
The distances must be perpendicular to the actual mirror line.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Tomorrow, explain why a point–image segment has its midpoint on and is perpendicular to the mirror.
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.