Reflection Across a Line: Geometric Transformation

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

Given a line \(L: ax + by + c = 0\) and a point \(P(x_0, y_0)\), the reflected point \(P'(x', y')\) is given by:

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

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

Special Cases

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

Point P Coordinates

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 \(L: 2x - y + 1 = 0\).

Solution

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

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

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 \(y = x\).

Solution

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 \(x - y = 0\)):

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

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

Example 3: Point on the mirror line

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

Solution

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

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.

5. Applications

6. 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 \(y = -\frac{1}{2}x + 2\).

Problem 3

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