Question

The image of a point after a reflection over the line y = –x is (7, –1). Find the coordinates of the preimage.

Answer 1

You know that your equation is y = -x. You know that your y for (7, -1) is -1. Therefore, you can solve for the preimage x by plugging -1 in for y:

y_image = -x_preimage -----> -1 = -x_preimage
x_preimage = 1

Conversely, since you flipped coordinates across a diagonal, you also know that:
x_image = -y_preimage

Therefore, you can solve for the preimage y by plugging -1 in for x:

x_image = -y_preimage -----> 7 = -y_preimage
y_preimage = -7

Your preimage coordinates are (1, -7).

Answer 2

Additionally, this is pretty easy to prove if you know linear algebra. A basis for a 2D Cartesian space is e1 = [1 0] and e2 = [0 1], where the contents inside the brackets represent a vector as [x_coordinate y_coordinate].

The identity matrix for a 2D Cartesian space is:
\[\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right] = \left[ \begin{matrix} \left(\begin{matrix}1 \\ 0\end{matrix}\right) & \left(\begin{matrix}0 \\ 1\end{matrix}\right) \end{matrix} \right] = \left[ \begin{matrix}e1 & e2 \end{matrix} \right]\]
Now, flipping across y = x means exchanging the order e1 and e2 in the matrix. Flipping acros y = -x means flipping across y = x and then multiplying the entire expression by -1. Therefore, the linear transformation for flipping across y = -x is:
\[-\left[\begin{matrix}e2 & e1 \end{matrix}\right] = -\left[\begin{matrix}\left(\begin{matrix}0 \\ 1\end{matrix}\right) & \left(\begin{matrix}1 \\ 0\end{matrix}\right) \end{matrix}\right] = \left[\begin{matrix}0 & -1 \\ -1 & 0\end{matrix}\right]\]
Now, you have a system of linear equations:
\[\left[\begin{matrix}0 & -1 \\ -1 & 0\end{matrix}\right]\left(\begin{matrix}x_{preimage} \\ y_{preimage} \end{matrix}\right) = \left(\begin{matrix}x_{image} \\ y_{image}\end{matrix}\right)\]
Using row reduction to solve this yields:
\[\left(\begin{matrix}x_{preimage} \\ y_{preimage}\end{matrix}\right) = \left(\begin{matrix}1 \\ -7\end{matrix}\right)\]

Both methods I have posted are actually the same thing. One is just a brute force equation representation. The other is just a linear algebra representation.