Reflection Matrix over Center Point

Negating Both Coordinates with a Matrix

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.

To find the $2 \times 2$ matrix for this transformation, start with unknown entries:

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

Reflection over the origin $O(0,0)$ negates both coordinates:

Applying the Reflection Matrix over the Origin

The matrix below turns a reflection over the origin into one multiplication. The examples apply it to single points and then to a whole triangle.

Reflecting Two Points

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

Solution:

Using the transformation matrix:

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

Both problems reflect over the origin. The second applies the reflection matrix to each vertex of a triangle.

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.

Solutions

Each solution below multiplies the reflection matrix by the coordinates of one point. Write the matrix first, then apply it to each vertex of the triangle so the image is built from its corners.

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:

   $$\begin{pmatrix} 2 & 5 & 3 \\ 2 & 2 & 5 \end{pmatrix}$$

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

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