Reflection Matrix over Arbitrary Point

Finding the Reflection Matrix over an Arbitrary Point

The image of a point $(x,y)$ reflected over the point $(a,b)$ is $(-x+2a, -y+2b)$ or $(2a-x, 2b-y)$.

The matrix operation associated with this transformation cannot be represented solely by a single $2 \times 2$ multiplication matrix, as it involves addition (translation) due to the center point $(a,b)$ not being the origin.

However, we can represent this transformation as a combination of matrix operations:

1. Translate the point $(x,y)$ so that the center of reflection $(a,b)$ effectively becomes the origin. This means we work with $(x-a, y-b)$.
2. Reflect this translated point over the origin using the matrix $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$.
3. Translate the result back by adding the coordinates of the center of reflection $(a,b)$.

Mathematically, if $(x', y')$ is the image of $(x,y)$:

This corresponds to the formula we are familiar with.

Matrix Operation for Reflection over a Point

The matrix operation associated with reflection over the point $P(a,b)$ for any point $(x,y)$ is:

Or, more precisely, it can be written as a combination:

The form presented as Property 4.11 in the book ($\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} + 2 \begin{pmatrix} a \\ b \end{pmatrix}$) is a simplification of $\begin{pmatrix} -x+2a \\ -y+2b \end{pmatrix}$.

Finding the Image of a Point

Determine the image of point $(2,3)$ by reflection over point $(1,1)$.

Alternative Solution:

Point $(x,y) = (2,3)$. Center $(a,b) = (1,1)$.

Visualization:

Exercises

1. Determine the image of point $(3,1)$ by reflection over point $(-2,-1)$.
2. A line passes through points $A(1,2)$ and $B(3,4)$. Determine the equation of the image line after reflection over point $C(0,1)$.

Key Answers

1. Point $(x,y) = (3,1)$. Center $(a,b) = (-2,-1)$.

   Using the formula $x' = 2a-x$ and $y' = 2b-y$:

   $$x' = 2(-2) - 3 = -4 - 3 = -7$$

   $$y' = 2(-1) - 1 = -2 - 1 = -3$$

   Its image is $(-7,-3)$. Or using matrix operations:

   $$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 3 \\ 1 \end{pmatrix} + 2 \begin{pmatrix} -2 \\ -1 \end{pmatrix} = \begin{pmatrix} -3 \\ -1 \end{pmatrix} + \begin{pmatrix} -4 \\ -2 \end{pmatrix} = \begin{pmatrix} -7 \\ -3 \end{pmatrix}$$

2. Center of reflection $C(0,1)$. ($a=0, b=1$)

   Image of point $A(1,2)$:

   $$x'_A = 2(0)-1 = -1$$

   $$y'_A = 2(1)-2 = 0$$

   So $A'(-1,0)$.

   Image of point $B(3,4)$:

   $$x'_B = 2(0)-3 = -3$$

   $$y'_B = 2(1)-4 = -2$$

   So $B'(-3,-2)$.

   The image line passes through $A'(-1,0)$ and $B'(-3,-2)$.

   Gradient $m' = \frac{-2 - 0}{-3 - (-1)} = \frac{-2}{-2} = 1$.

   Equation of the line:

   $$y - y'_A = m'(x - x'_A)$$

   $$y - 0 = 1(x - (-1))$$

   $$y = x + 1$$

   or

   $$x - y + 1 = 0$$