How do you find the reflection of a point across a line in geometry?

Given a line ax + by + c = 0 and a point P(x0, y0), the reflected point P'(x', y') is calculated using the formulas x' = x0 - 2a((ax0 + by0 + c)/(a^2 + b^2)) and y' = y0 - 2b((ax0 + by0 + c)/(a^2 + b^2)).

Reflection Across a Line – Formulas & Examples

In geometry, a reflection is a transformation that maps a point to its mirror image across a line. The line is called the mirror line or axis of reflection. Reflection preserves distances and angles, making it an isometry (distance-preserving transformation).

Key properties: The mirror line is the perpendicular bisector of the segment joining a point and its reflected image. Points on the mirror line remain fixed.

1. Mathematical Formula for Reflection

The reflection \( P'(x', y') \) of a point \( P(x_0, y_0) \) across the line \( L: ax + by + c = 0 \) is calculated as:

\[\begin{aligned} x' &= x_0 - 2a \dfrac{ax_0 + by_0 + c}{a^2 + b^2} \\ y' &= y_0 - 2b \dfrac{ax_0 + by_0 + c}{a^2 + b^2} \end{aligned}\]

Where \(d = \dfrac{ax_0 + by_0 + c}{\sqrt{a^2 + b^2}}\) is the signed distance from point \(P\) to the line.

Special Cases

Reflection across the x-axis (\(y = 0\)): \((x, y) \to (x, -y)\)
Reflection across the y-axis (\(x = 0\)): \((x, y) \to (-x, y)\)
Reflection across the line \(y = x\): \((x, y) \to (y, x)\)
Reflection across the line \(y = -x\): \((x, y) \to (-y, -x)\)

2. Interactive Visualization

Mirror Line: \(y = mx + b\)

Adjust the line parameters or drag the point to see its reflection in real time.

Line Slope & Intercept
m:
b:

Point P Coordinates
x:
y:

Actions

Blue: original point | Red: reflected point | Dashed: mirror line

3. Worked Examples

Example 1: Reflection across the line \(y = 2x + 1\)

Find the reflection of point \(P(4, 3)\) across the line \(L: 2x - y + 1 = 0\).

Step 1: Identify coefficients: \(a = 2\), \(b = -1\), \(c = 1\).

Step 2: Compute \(d = \dfrac{ax_0 + by_0 + c}{a^2 + b^2} = \dfrac{2(4) + (-1)(3) + 1}{2^2 + (-1)^2} = \dfrac{8 - 3 + 1}{4 + 1} = \dfrac{6}{5} = 1.2\).

Step 3: Apply reflection formula:

\[\begin{aligned} x' &= x_0 - 2a d = 4 - 2(2)(1.2) = 4 - 4.8 = -0.8 \\ y' &= y_0 - 2b d = 3 - 2(-1)(1.2) = 3 + 2.4 = 5.4 \end{aligned}\]

Thus, the reflected point is \(P'(-0.8, 5.4)\).

Example 2: Reflection across the line \(y = x\)

Reflect \(Q(5, -2)\) across \(y = x\).

For \(y = x\), the transformation simply swaps coordinates: \((x, y) \to (y, x)\).

Therefore, \(Q'(-2, 5)\).

We can verify using the formula with \(a = 1\), \(b = -1\), \(c = 0\) (since \(x - y = 0\)):

\[d = \dfrac{5 - (-2)}{1^2 + (-1)^2} = \dfrac{7}{2} = 3.5\] \[x' = 5 - 2(1)(3.5) = 5 - 7 = -2\] \[y' = -2 - 2(-1)(3.5) = -2 + 7 = 5\]

Matching the result \((-2, 5)\).

Example 3: Point on the mirror line

Reflect \(R(3, 4)\) across the line \(3x - 4y + 7 = 0\).

Check if \(R\) lies on the line: \(3(3) - 4(4) + 7 = 9 - 16 + 7 = 0\).

Since the point is on the line, it is fixed: \(R' = R(3, 4)\).

The formula confirms: \(d = 0\) → \(x' = 3\), \(y' = 4\).

4. Reflection as an Isometry & Applications

Reflection preserves distances: if \(P\) and \(Q\) are two points, and \(P'\) and \(Q'\) are their reflections, then:

\[\text{dist}(P, Q) = \text{dist}(P', Q')\]

It also preserves angles, making it a rigid motion. Reflection reverses orientation (handedness) — it's an improper rotation.

Applications

Computer Graphics: Mirror effects, symmetry operations.
Physics: Law of reflection (angle of incidence equals angle of reflection).
Geometry: Constructing symmetric figures, solving optimization problems (shortest path).

5. Practice Problems

Problem 1

Find the reflection of \((-2, 5)\) across the line \(x = 3\).

Hint: vertical line — the y-coordinate stays the same.

Problem 2

Reflect \((1, -3)\) across the line \(y = -\dfrac{1}{2}x + 2\).

Problem 3

Show that reflecting a point twice across the same line returns the original point.