Ever wondered why a picture in a mirror looks flipped? That's reflection, and in coordinate geometry you can flip points on a graph just the same.
In Simple Words: Reflection is like holding a mirror to a point on the graph. The point you see in the mirror is called the image, and the original point is the pre‑image. The mirror line tells you where to flip.
What is Reflection in Coordinate Geometry?
Reflection means creating a mirror image of a shape across a straight line called the axis of reflection. The original shape is the pre‑image, and its copy is the image. Think of the line as a river; everything on one side has a twin on the opposite bank at the same distance.
Common Reflection Lines and Their Formulas
In the ICSE board, you’ll meet four favourite lines. Each one has a neat shortcut to find the image (x', y') of any point (x, y).
| Reflection Line | How to get (x', y') |
|---|---|
| y‑axis (x = 0) | (-x, y) – just change the sign of x |
| x‑axis (y = 0) | (x, -y) – just change the sign of y |
| line y = x | (y, x) – swap the coordinates |
| line y = -x | (-y, -x) – swap and change both signs |
These shortcuts save you minutes in an exam.
Step‑by‑Step Method for Any Arbitrary Line
When the mirror line isn’t one of the four easy ones, you follow a short process. It’s like building a tiny bridge from the point to the line, walking across, and landing on the other side.
Worked Example 1: Reflect (3, 4) across the line y = 2x – 1
1. Write the line in the form Ax + By + C = 0. Here, y – 2x + 1 = 0, so A = –2, B = 1, C = 1.
2. Use the formula for the image of (x, y) across Ax + By + C = 0:
x' = x – 2A (Ax + By + C) / (A² + B²)
y' = y – 2B (Ax + By + C) / (A² + B²)
Plug in the numbers:
- Ax + By + C = (–2)(3) + (1)(4) + 1 = –6 + 4 + 1 = –1
- A² + B² = (–2)² + 1² = 4 + 1 = 5
Now compute:
- x' = 3 – 2(–2)(–1)/5 = 3 – (4/5) = 2.2
- y' = 4 – 2(1)(–1)/5 = 4 + (2/5) = 4.4
So the image is (2.2, 4.4). Check: the midpoint of (3, 4) and (2.2, 4.4) lies on the line, and the segment joining them is perpendicular – both conditions for a correct reflection.
Worked Example 2: Reflect the triangle with vertices A(1,2), B(4,2), C(4,5) across the line x = 3
Since x = 3 is a vertical line, just flip the x‑coordinate: x' = 2·3 – x = 6 – x.
- A' = (5,2)
- B' = (2,2)
- C' = (2,5)
The reflected triangle is a mirror image on the other side of the line x = 3.
Quick Summary
- Identify the mirror line.
- If it’s y‑axis, x‑axis, y = x, or y = –x, use the shortcut table.
- For any other line, rewrite it as Ax + By + C = 0 and apply the general formula.
- Always verify by checking (a) the midpoint lies on the line and (b) the joining segment is perpendicular.
📝 Likely Exam Questions
- Question: Find the image of the point (‑2, 5) when reflected in the line y = –x.
Answer: Swap coordinates and change both signs → (‑5, 2). - Question: A point P(7, ‑3) is reflected in the line 3x + 4y – 12 = 0. Find the coordinates of the image P'.
Answer: Compute Ax+By+C = 3·7 + 4·(‑3) – 12 = 21 – 12 – 12 = ‑3. A²+B² = 9+16=25.
x' = 7 – 2·3·(‑3)/25 = 7 + 18/25 = 7.72.
y' = ‑3 – 2·4·(‑3)/25 = ‑3 + 24/25 = ‑2.04 (approx). - Question: The line y = 2x is the axis of reflection. What is the image of the line 3x + y – 6 = 0?
Answer: Replace (x, y) by (y, x) in the original equation → 3y + x – 6 = 0, or x + 3y – 6 = 0. - Question: A triangle has vertices (1,1), (4,1), (4,4). Find its image after reflection in the x‑axis.
Answer: Change the sign of y‑coordinates → (1,‑1), (4,‑1), (4,‑4). - Question: Explain how you would check whether a reflected point is correct.
Answer: Find the midpoint of the original and image; it must satisfy the equation of the mirror line. Also, the line joining the two points must be perpendicular to the mirror line (their slopes multiply to –1).