Image of a Point in a Plane

Image (reflection) of P(x₁,y₁,z₁) in plane ax+by+cz+d=0. The image P′ is such that the foot of perpendicular is the midpoint of PP′. The formula follows by doubling the foot-of-perpendicular displacement.

Let $P(x_1,y_1,z_1)$ have image $P'(x',y',z')$ in the plane $ax+by+cz+d=0$.

By definition of reflection: the foot of the perpendicular from $P$ to the plane is the midpoint $M$ of $PP'$.

From the foot derivation, $M = (x_1+ak, y_1+bk, z_1+ck)$ where $k = -\dfrac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$.

Since $M$ is the midpoint of $PP'$:

$$x' = 2x_M - x_1 = x_1+2ak$$ $$y' = 2y_M - y_1 = y_1+2bk$$ $$z' = 2z_M - z_1 = z_1+2ck$$

Writing in the symmetric form:

$$\frac{x'-x_1}{a} = \frac{y'-y_1}{b} = \frac{z'-z_1}{c} = 2k = -\frac{2(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}$$

Note: The image formula has a factor of 2 compared to the foot formula — because the foot is halfway, and the image is the full reflection.

Image in a line (2D analogue): Exactly the same logic applies — foot of perpendicular is the midpoint, and the image is the symmetric point.