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Geometric Transformations

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 rotation about the origin.

Reflection Matrix over Center PointLearn reflection matrices over the origin. Learn point transformations with worked calculations and visual examples.

On this page

Finding the Reflection Matrix over the Origin
- Reflection Matrix over the Origin
Application of Reflection Matrix over the Origin
Finding the Image of Points
Exercises
- Key Answers

Comments

180^\circ

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:

\begin{pmatrix} r & s \\ t & u \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -x \\ -y \end{pmatrix}

From matrix multiplication, we can write:

\begin{pmatrix} rx + sy \\ tx + uy \end{pmatrix} = \begin{pmatrix} -1x + 0y \\ 0x - 1y \end{pmatrix}

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:

\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}

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 $A(-1,1)$ :

\begin{pmatrix} x'_A \\ y'_A \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} -1 \\ 1 \end{pmatrix} = \begin{pmatrix} (-1)(-1) + (0)(1) \\ (0)(-1) + (-1)(1) \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \end{pmatrix}

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

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

\begin{pmatrix} x'_B \\ y'_B \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} (-1)(3) + (0)(-2) \\ (0)(3) + (-1)(-2) \end{pmatrix} = \begin{pmatrix} -3 \\ 2 \end{pmatrix}

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

Reflection of Points

A

and

B

over the Origin

Visualization of reflecting point

A(-1,1)

to

A'(1,-1)

and

B(3,-2)

to

B'(-3,2)

over the origin

O(0,0)

.

Exercises

Determine the images of points $A(1,-7)$ and $B(-7,-2)$ when reflected over the origin!
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

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 $B'(7,2)$ .
Matrix of PQR vertices: $\begin{pmatrix} 2 & 5 & 3 \\ 2 & 2 & 5 \end{pmatrix}$ .

Reflection Matrix over Arbitrary Point

Reflection over a Line

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