Where do points go in a reflection?
Pick the line of reflection and type the points of your figure. The reflection calculator flips each point to the other side of the line, the same distance away, and shows the rule and the working.
- Image points
- A′(3, −2); B′(5, −2); C′(4, −6)
The image points are A′(3, −2); B′(5, −2); C′(4, −6).
- Rule
- (x, y) → (x, −y)
- Working
- rule (x, y) → (x, −y); A(3, 2) → A′(3, −2); B(5, 2) → B′(5, −2); C(4, 6) → C′(4, −6)
Image points: A′(3, −2); B′(5, −2); C′(4, −6). The image points are A′(3, −2); B′(5, −2); C′(4, −6).
How it is worked out
How to calculate
Reflects points in the coordinate plane across the x-axis, the y-axis, the origin, y = x, y = −x, a line x = h or y = k, or any line y = mx + b, with the rule and the working.
Example with the default inputs (Reflect across x-axis, Points [x 3, y 2; x 5, y 2; x 4, y 6]): The image points are A′(3, −2); B′(5, −2); C′(4, −6).
Method: x-axis (x, −y); y-axis (−x, y); origin (−x, −y); y = x (y, x); y = −x (−y, −x); x = h (2h − x, y); y = k (x, 2k − y); y = mx + b: d = (x + m(y − b)) ÷ (1 + m²), (2d − x, 2md − y + 2b).
- Typed decimals are read exactly, so each image coordinate is the exact value, rounded once to 10 significant figures.
- A reflection keeps every length and angle; it flips the figure over the line.
Worked examples
Each example is checked against the calculator on every build.
- Reflect across x-axis, Points 3 2 gives Image points A′(3, −2), Rule (x, y) → (x, −y).Source: CK-12 Foundation, Geometry, 8.14 Rules for Reflections ((x, y) → (x, −y) over the x-axis, (−x, y) over the y-axis, (y, x) over y = x, (−y, −x) over y = −x; (3, 2) over the x-axis is (3, −2), over y = x is (2, 3)), https://k12.libretexts.org/Bookshelves/Mathematics/Geometry/08%3A_Rigid_Transformations/8.14%3A_Rules_for_Reflections (retrieved 2026-10-05)
- Reflect across Line y = x, Points 3 2 gives Image points A′(2, 3).Source: CK-12 Foundation, Geometry, 8.14 Rules for Reflections ((x, y) → (x, −y) over the x-axis, (−x, y) over the y-axis, (y, x) over y = x, (−y, −x) over y = −x; (3, 2) over the x-axis is (3, −2), over y = x is (2, 3)), https://k12.libretexts.org/Bookshelves/Mathematics/Geometry/08%3A_Rigid_Transformations/8.14%3A_Rules_for_Reflections (retrieved 2026-10-05)
- Reflect across y-axis, Points 3 2; -1.5 4 gives Image points A′(−3, 2); B′(1.5, 4).Source: OpenStax, Algebra and Trigonometry 2e, §3.5 Transformation of Functions (−f(x) reflects about the x-axis, f(−x) about the y-axis), https://openstax.org/books/algebra-and-trigonometry-2e/pages/3-5-transformation-of-functions (retrieved 2026-10-05)
- Reflect across Vertical line x = h, h 2, Points 5 -1 gives Image points A′(−1, −1).
- Reflect across Line y = mx + b, Slope m 2, Intercept b 1, Points 3 2 gives Image points A′(−1, 4).
How it works
A reflection sends each point P to the point P′ on the other side of the line, so that the line is the perpendicular bisector of PP′. For each choice:
| Reflect across | Image of (x, y) |
|---|---|
| x-axis | (x, −y) |
| y-axis | (−x, y) |
| the origin (0, 0) | (−x, −y) |
| y = x | (y, x) |
| y = −x | (−y, −x) |
| vertical line x = h | (2h − x, y) |
| horizontal line y = k | (x, 2k − y) |
| line y = mx + b | (2d − x, 2md − y + 2b), with d = (x + m(y − b)) ÷ (1 + m²) |
Why the last rule works. The point on y = mx + b closest to P = (x, y) is F = (d, md + b), where d = (x + m(y − b)) ÷ (1 + m²): the line from P to F is at a right angle to the line's direction (1, m). F is the midpoint of P and P′, so P′ = 2F − P = (2d − x, 2md + 2b − y).
Rules
- Every coordinate, h, k, m and b is from −10¹² to 10¹². Leave h, k, m or b at its default when the chosen line does not use it.
- Up to 12 points, named A, B, C and so on.
Output format. Typed decimals are read exactly, so each image coordinate is computed exactly and rounded once to 10 significant figures, with the true minus sign (−).
Worked examples by hand
Over the x-axis. (3, 2) → (3, −2).
Over y = x. (3, 2) → (2, 3).
Over the y-axis. (3, 2) → (−3, 2) and (−1.5, 4) → (1.5, 4).
Over x = 2. (5, −1) → (2 × 2 − 5, −1) = (−1, −1).
Over y = 2x + 1. For (3, 2): d = (3 + 2 × (2 − 1)) ÷ (1 + 4) = 5 ÷ 5 = 1. The image is (2 × 1 − 3, 2 × 2 × 1 − 2 + 2 × 1) = (−1, 4). Check: the midpoint of (3, 2) and (−1, 4) is (1, 3), which is on y = 2x + 1.
Other questions people ask
How do I reflect a point over the x-axis?
Keep x and change the sign of y: (x, y) → (x, −y). The point (3, 2) reflected over the x-axis is (3, −2).
How do I reflect a point over the y-axis?
Change the sign of x and keep y: (x, y) → (−x, y). The point (3, 2) goes to (−3, 2).
What is the rule for a reflection over y = x?
Swap the coordinates: (x, y) → (y, x). The point (3, 2) goes to (2, 3). Over y = −x, swap them and change both signs: (x, y) → (−y, −x).
How do I reflect over a line such as x = 2 or y = −1?
Over the vertical line x = h, the image is (2h − x, y); over the horizontal line y = k, it is (x, 2k − y). The point (5, −1) reflected over x = 2 is (4 − 5, −1) = (−1, −1).
How do I reflect a point over any line y = mx + b?
Find d = (x + m(y − b)) ÷ (1 + m²), the x coordinate of the point on the line closest to (x, y). The image is (2d − x, 2md − y + 2b). For (3, 2) and y = 2x + 1, d = 1 and the image is (−1, 4).
Is a reflection over the origin a reflection over a line?
No. Reflecting through the origin sends (x, y) to (−x, −y), the same as a 180° rotation about the origin. It is listed because textbooks call it a point reflection.