Reflection Matrix over Center Point

Finding the Reflection Matrix over the Origin

Reflecting a point $(x,y)$ over the origin $O(0,0)$ results in the image $(-x,-y)$. This is equivalent to a $180^\circ$ rotation about the origin.

Now, we will find the $2 \times 2$ matrix, let's say $\begin{pmatrix} r & s \\ t & u \end{pmatrix}$, that represents this transformation.

We want to find $r, s, t, u$ such that:

From matrix multiplication, we can write:

By equating the corresponding coefficients, we get:

• For the first row: $rx + sy = -1x + 0y$. This means $r = -1$ and $s = 0$.
• For the second row: $tx + uy = 0x - 1y$. This means $t = 0$ and $u = -1$.

Reflection Matrix over the Origin

The matrix associated with reflection over the origin $O(0,0)$ is:

Application of Reflection Matrix over the Origin

Finding the Image of Points

Determine the images of points $A(-1,1)$ and $B(3,-2)$ when reflected over the origin!

Alternative Solution:

Using the transformation matrix $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$:

For point $A(-1,1)$:

The image of point A is $A'(1,-1)$.

For point $B(3,-2)$:

The image of point B is $B'(-3,2)$.

Exercises

1. Determine the images of points $A(1,-7)$ and $B(-7,-2)$ when reflected over the origin!
2. A triangle $PQR$ has vertices $P(2,2)$, $Q(5,2)$, and $R(3,5)$. Determine the coordinates of the image triangle $P'Q'R'$ after reflection over the origin using matrix multiplication.

Key Answers

1. The reflection matrix over the origin is: $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$.

   For $A(1,-7)$:

   $$\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 1 \\ -7 \end{pmatrix} = \begin{pmatrix} -1 \\ 7 \end{pmatrix}$$

   Image $A'(-1,7)$. For $B(-7,-2)$:

   $$\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} -7 \\ -2 \end{pmatrix} = \begin{pmatrix} 7 \\ 2 \end{pmatrix}$$

   Image $B'(7,2)$.

2. Matrix of PQR vertices: $\begin{pmatrix} 2 & 5 & 3 \\ 2 & 2 & 5 \end{pmatrix}$.

   $$\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 2 & 5 & 3 \\ 2 & 2 & 5 \end{pmatrix} = \begin{pmatrix} -2 & -5 & -3 \\ -2 & -2 & -5 \end{pmatrix}$$

   Image: $P'(-2,-2)$, $Q'(-5,-2)$, $R'(-3,-5)$.