On the reflection of a point over a line

Did some brain-storming to come up with this (even if it is probably already done) formula to find the reflection P'(x₂, y₂) of a point P(x₁, y₁) over a line:

Ax + By + C = 0 --- (1)

 Idea: I will use two equations in x₂ and y₂, containing constants A, B, C, x₁ and y₁, to derive the value of x₂ and y₂ in terms of the said constants. For the first equation, I shall take the help of the equation of the straight line passing through P and P'. For the second equation, I shall use the midpoint formula for the point M(a, b) that lies on line (1) and lies between P and P'.

Ok, let’s look at the figure first. Suffice to say, it represents the reflection of any point over a general straight line.

PP' is a line that is perpendicular to (1)

So the equation of PP’ would be:

Bx – Ay + K = 0 --- (2)

where K is a constant

[reason: the slope of line (1) is found to be –A/B, and since line (2) is perpendicular to line (1), its slope must be B/A (to give a product of –1), which gives us equation (2)]

As P(x₁, y₁) lies on line (2),

Bx₁ – Ay₁ + K = 0

K = Ay₁ – Bx₁

Thus, line (2) can be written as:

Bx – Ay + (Ay₁ – Bx₁) = 0 --- (2)

As P'(x₂, y₂) also lies on this line,

Bx₂ – Ay₂ + (Ay₁ – Bx₁) = 0

=> Ay₂ = Bx₂ + (Ay₁ – Bx₁)

=> y₂ = [Bx₂ + (Ay₁ – Bx₁)]/A --- (A)

Now let us look at the point M(a, b) which lies on line (1) and is half-way between P and P'

(As P' is the reflection of P over line (1), PM = MP')

By the mid-point formula,

(x₁ + x₂)/2 = a

(y₁ + y₂)/2 = b

(a, b) lies on line (1), so:

Aa + Bb + C = 0

=> A(x₁ + x₂)/2 + B(y₁ + y₂)/2 + C = 0

=> [Ax₁ + Ax₂ + By₁ + By₂ + 2C]/2 = 0

=> Ax₂ + By₂ + [Ax₁ + By₁ + 2C] = 0

=> By₂ = -Ax₂ – (Ax₁ + By₁ + 2C)

=> y₂ = [-Ax₂ – Ax₁ – By₁ – 2C]/B --- (B)

Thus from (A) and (B),

Bx₂ + (Ay₁ – Bx₁)     =      -Ax₂ – Ax₁ – By₁ – 2C

A                                         B

B²x₂ + B(Ay₁ – Bx₁) = -A²x₂ – A²x₁ – ABy₁ – 2AC

(A² + B²)x₂ = - ABy₁ + B²x₁ – A²x₁ – ABy₁ – 2AC

x₂ = (B² – A²)x₁ – 2ABy₁ – 2AC

                   (A² + B²)

Therefore, from (A),

y₂ = B[(B² – A²)x₁ – 2ABy₁ – 2AC]/(A² + B²) + Ay₁ – Bx₁

                                                A

= B³x₁ – A²Bx₁ – 2AB²y₁ – 2ABC + A³y₁ + AB²y₁ – A²Bx₁ – B³x₁

                                    A(A² + B²)

= A³y₁ – 2A²Bx₁ – AB²y₁ – 2ABC

                        A(A² + B²)

= A²y₁ – 2ABx₁ – B²y₁ – 2BC

                        (A² + B²)

y₂ = (A² – B²)y₁ – 2ABx₁ – 2BC

                   (A² + B²)

Let’s try this formula for a certain case, when a point is reflected on the line:

y = -x

i.e. x + y = 0

Then, A = 1, B = 1, C = 0

x₂ = 0 – 2y₁ – 0   = -y₁

            1 + 1

y₂ = 0 – 2x₁ – 0   = -x₁

            1 + 1