Reflection of a Point in a Line

Image (reflection) of point (x₁, y₁) in the line ax + by + c = 0.

The image $P'$ of point $P(x_1, y_1)$ in line $L: ax + by + c = 0$ is the point such that $L$ is the perpendicular bisector of $PP'$.

From the foot of perpendicular derivation, the foot $N$ corresponds to parameter:

$$t = -\frac{ax_1 + by_1 + c}{a^2 + b^2}$$

Since $N$ is the midpoint of $PP'$, the image $P' = P + 2\overrightarrow{PN}$. The displacement $\overrightarrow{PN} = (at, bt)$, so:

$$P' = (x_1 + 2at,\; y_1 + 2bt)$$

Substituting $t$:

$$\boxed{\frac{x - x_1}{a} = \frac{y - y_1}{b} = -\frac{2(ax_1 + by_1 + c)}{a^2 + b^2}}$$

The foot uses parameter $t$ once; the image uses $2t$. This is the only difference between the two results.

> > | Mirror line | $a$ | $b$ | $c$ | Image of $(x_1, y_1)$ | > |---|---|---|---|---| > | $x$-axis ($y = 0$) | $0$ | $1$ | $0$ | $(x_1,\ -y_1)$ | > | $y$-axis ($x = 0$) | $1$ | $0$ | $0$ | $(-x_1,\ y_1)$ | > | $y = x$ | $1$ | $-1$ | $0$ | $(y_1,\ x_1)$ | > | $y = -x$ | $1$ | $1$ | $0$ | $(-y_1,\ -x_1)$ | > | Origin | — | — | — | $(-x_1,\ -y_1)$ | > | $x = k$ | $1$ | $0$ | $-k$ | $(2k - x_1,\ y_1)$ | > | $y = k$ | $0$ | $1$ | $-k$ | $(x_1,\ 2k - y_1)$ | > > The origin is a point, not a line — its reflection is a rotation by $180°$, equivalent to reflecting twice (in $x$-axis then $y$-axis).