Reflections of Points, Lines and Planes
The reflection of a point P in a point S, a line or a plane is the point P′ such that the object bisects the segment PP′ perpendicularly; in formulas P′ = 2S − P for a point, P′ = 2F − P for a line, and P′ = P − 2(f(P)/|n|2) n for the plane n · X + D = 0, where F is the foot of the perpendicular. In brief: the reflection of (2, 3) in the line 3x + 4y − 10 = 0 is (2/25, 11/25), and the reflection of (1, 2, 3) in the plane 2x − 2y + z − 9 = 0 is (41/9, −14/9, 43/9).
Reflections are the simplest isometries that change orientation, and they are also used to solve shortest-path problems. This guide shows how to reflect points in a point, a line (in the plane and in space) and a plane, and then how to reflect whole lines and planes, with worked examples and a summary table.
How do you reflect a point in a point?
The center S is the midpoint of PP′, so P′ = 2S − P.
Example 1: central symmetry
P(1, 2, 3) and S(0, 1, −1): P′ = (0, 2, −2) − (1, 2, 3) = (−1, 0, −5).
How do you reflect a point in a line?
Find the foot F of the perpendicular from P to the line, and then P′ = 2F − P. For special lines there are shortcuts: in the x-axis (a, b) goes to (a, −b); in the y-axis to (−a, b); in the line y = x to (b, a).
Example 2: a line in the plane
P(2, 3) in 3x + 4y − 10 = 0: the foot is F = (26/25, 43/25) (from the previous article). P′ = (52/25 − 2, 86/25 − 3) = (2/25, 11/25). A direct formula: P′ = P − 2(f(P)/(A2 + B2))(A, B) = (2, 3) − (16/25)(3, 4) = (2/25, 11/25) ✓.
Example 3: a line in space
P(0, 0, 0) in the line (1, 2, 3) + t(3, −2, 2): the foot is (2/17, 44/17, 41/17), so P′ = (4/17, 88/17, 82/17). The distance |PP′| equals twice the distance from P to the line, 2 · 3.54 ≈ 7.08: √(16 + 7744 + 6724)/17 = √14484/17 ≈ 7.08 ✓.
Example 4: a line through the origin
The reflection in the line y = 2x, direction (1, 2): the reflection of P(a, b) is P′ = 2((P · d)/(d · d)) d − P. For P = (5, 0): P · d = 5, d · d = 5, so P′ = 2(1, 2) − (5, 0) = (−3, 4). Check: |P′| = 5 = |P| ✓.
How do you reflect a point in a plane?
With f(X) = n · X + D, the reflection is P′ = P − 2(f(P)/|n|2) n. The foot F is the midpoint of PP′.
Example 5: a plane reflection
P(1, 2, 3) in 2x − 2y + z − 9 = 0: f(P) = −8, |n|2 = 9, P′ = (1, 2, 3) + (16/9)(2, −2, 1) = (41/9, −14/9, 43/9). The midpoint of PP′ is ((9 + 41)/18, (18 − 14)/18, (27 + 43)/18) = (25/9, 2/9, 35/9), the foot found in the previous article ✓.
Example 6: reflection in the plane x = z
With n = (1, 0, −1) and D = 0: P′ = P − (x − z)(1, 0, −1) = (z, y, x). The mirror x = z exchanges the coordinates x and z.
How do you reflect a line or a plane?
An isometry maps lines to lines and planes to planes, so it is enough to reflect points (two points of a line, three points of a plane) or a point together with the direction. The direction vector d of a line is reflected in a plane by d′ = d − 2((n · d)/|n|2) n.
Example 7: a line in a plane
The line L through the origin with d = (1, 0, 0) and the mirror x + y + z = 1 (n = (1, 1, 1), |n|2 = 3). L meets the mirror at (1, 0, 0), which stays fixed. The direction becomes d′ = (1, 0, 0) − (2/3)(1, 1, 1) = (1/3, −2/3, −2/3), parallel to (1, −2, −2). The reflected line: (1, 0, 0) + s(1, −2, −2). Check with the reflected origin: the origin reflects to (2/3, 2/3, 2/3), and (2/3, 2/3, 2/3) − (1, 0, 0) = (−1/3, 2/3, 2/3) is parallel to (1, −2, −2) ✓.
Example 8: a plane in a plane
The plane x = 1 in the mirror x = z. Since the mirror exchanges x and z (Example 6), the image is z = 1.
Example 9: a plane in a plane, checked by points
The plane 6x + 3y + 2z = 6 in the mirror x = z becomes 6z + 3y + 2x = 6, that is 2x + 3y + 6z = 6, by exchanging x and z.
Where is reflection used for shortest paths?
The shortest path from A to B that touches a line must be found by reflecting one of the points. If A and B are on the same side of the line, reflect B to B′; the shortest path is the segment AB′, which meets the line at the touching point, because |AX| + |XB| = |AX| + |XB′| ≥ |AB′|. This explains the law of reflection of light: the angle of incidence equals the angle of reflection.
Example 10: a shortest path
A(1, 4) and B(7, 2); the path must touch the x-axis. Reflect B to B′(7, −2). The length of the shortest path is |AB′| = √(36 + 36) = 6√2 ≈ 8.49. The touching point is where AB′ crosses y = 0: from (1, 4) moving by (6, −6) t, y = 0 at t = 2/3, so x = 5; the point is (5, 0).
Table of reflection formulas
| Mirror | Formula for P′ | Example | Result |
|---|---|---|---|
| point S | 2S − P | (1, 2, 3) in (0, 1, −1) | (−1, 0, −5) |
| x-axis | (a, −b) | (2, 3) | (2, −3) |
| line y = x | (b, a) | (2, 3) | (3, 2) |
| line Ax + By + C = 0 | P − 2f(P)/(A2 + B2) · (A, B) | (2, 3), 3x + 4y − 10 = 0 | (2/25, 11/25) |
| line in space | 2F − P | (0, 0, 0) | (4/17, 88/17, 82/17) |
| plane n · X + D = 0 | P − 2f(P)/|n|2 · n | (1, 2, 3), 2x − 2y + z − 9 = 0 | (41/9, −14/9, 43/9) |
Where does the idea come from?
The law of reflection and the minimal path property were known to the Greeks: Hero of Alexandria showed in the first century that the path of reflected light is shortest. Reflections as transformations of the plane are classified among the isometries.
Which mistakes are common?
Stopping at the foot
The foot F is the midpoint, not the image. Double the step: P′ = 2F − P.
Wrong sign of f(P)
The formula P′ = P − 2(f(P)/|n|2) n keeps its sign automatically; do not take an absolute value of f(P).
Reflecting only one point of a line
A line needs two points (or a point and a reflected direction) to be reflected.
Reflecting both points in the shortest path problem
Reflect only one of the points. Reflecting both gives the same length but loses the geometric meaning.