Reflection of a Line in Another Line

Image of line L₁ in mirror line L₂. Method: reflect any two points on L₁ using the point-reflection formula, then find the line through their images.

To reflect line $L_1$ in mirror line $L_2$, note that every point on the image line $L_1'$ is the reflection of some point on $L_1$ in $L_2$. A line is determined by two points, so reflecting any two points on $L_1$ and joining their images gives $L_1'$.

Setup

Let $L_1: a_1 x + b_1 y + c_1 = 0$ and mirror $L_2: a_2 x + b_2 y + c_2 = 0$.

Pick two convenient points on $L_1$ — typically where $L_1$ cuts the axes, or any two points easy to compute with.

Reflecting Each Point

For each point $P(x_1, y_1)$ on $L_1$, its image $P'(x', y')$ in $L_2$ satisfies:

$$\frac{x' - x_1}{a_2} = \frac{y' - y_1}{b_2} = -\frac{2(a_2 x_1 + b_2 y_1 + c_2)}{a_2^2 + b_2^2}$$

Let $k = -\dfrac{2(a_2 x_1 + b_2 y_1 + c_2)}{a_2^2 + b_2^2}$. Then:

$$x' = x_1 + a_2 k, \qquad y' = y_1 + b_2 k$$

Getting the Image Line

Apply this to two points $P_1, P_2$ on $L_1$ to get $P_1', P_2'$. The image line $L_1'$ passes through both.

If $P_1' = (x_1', y_1')$ and $P_2' = (x_2', y_2')$, the image line is:

$$\frac{x - x_1'}{x_2' - x_1'} = \frac{y - y_1'}{y_2' - y_1'}$$

or equivalently by the two-point form.

Special Case: $L_1 \parallel L_2$

When $L_1 \parallel L_2$, the image $L_1'$ is also parallel to both, on the opposite side of $L_2$ at the same distance. If $d$ is the distance between $L_1$ and $L_2$, then $L_1'$ is at distance $2d$ from $L_1$.

Special Case: $L_1$ passes through the intersection of $L_2$ with another line

If $L_1$ and $L_2$ intersect at point $Q$, then $Q$ lies on $L_2$ and is its own reflection. So $Q$ also lies on $L_1'$. Only one more reflected point is needed to determine $L_1'$.

> > Take $P_1 = (3, 0)$ on $L_1$. For $L_2: a_2=1, b_2=-1, c_2=-1$: > > $$ > k = -\frac{2(3 - 0 - 1)}{1 + 1} = -\frac{4}{2} = -2 > $$ > > $$ > P_1' = (3 + 1\cdot(-2),\ 0 + (-1)\cdot(-2)) = (1,\ 2) > $$ > > Take $P_2 = (0, 3)$ on $L_1$: > > $$ > k = -\frac{2(0 - 3 - 1)}{2} = -\frac{-8}{2} = 4 > $$ > > $$ > P_2' = (0 + 4,\ 3 + (-1)\cdot 4) = (4,\ -1) > $$ > > Line through $(1, 2)$ and $(4, -1)$: slope $= \dfrac{-1-2}{4-1} = -1$. > > $$ > y - 2 = -1(x - 1) \implies x + y = 3 > $$ > > Here $L_1' = L_1$ because $L_1$ is perpendicular to $L_2$ at their intersection — the line reflects onto itself.