Algorithm for reflecting a point across a line

Question

Given a point (x1, y1) and an equation for a line (y=mx+c), I need some pseudocode for determining the point (x2, y2) that is a reflection of the first point across the line

Answer 1

With reference to the fig in here.

We want to find the reflection of the point A(p,q) to line L1,eqn y = m*x + c. Say reflected point is A'(p',q')

Suppose, The line joining the points A and A' is L2 with eqn: y= m'*x + c' L1 & L2 intersect at M(a,b)

The algorithm for finding the reflection of the point is as follows: 1) Find slope of L2 is = -1/m , as L1 and L2 are perpendicular 2) Using the m' and A(x,y) find c' using eqn of L2 3) Find the intersection point 'M' of L1 anSd L2 4) As now we have coordinate of A and M so coordinate of A' can be easily obtained using the relation [ A(p,q)+A'(p',q') ]/2 = M(a,b)

I haven't checked the following code but the crude form of code in the FORTRAN is

SUBROUTINE REFLECTION(R,p,q)

IMPLICIT NONE

REAL,INTENT(IN)     ::  p,q

REAL, INTENT(OUT)   ::  R(2)

REAL                ::  M1,M2,C1,C2,a,b

M2=-1./M1                       ! CALCULATE THE SLOPE OF THE LINE L2 

C2=S(3,1)-M2*S(3,2)             ! CALCULATE THE 'C' OF THE LINE L2  

q= (M2*C1-M1*C2)/(M2-M1)        ! CALCULATE THE MID POINT O

p= (q-C1)/M1

R(1)=2*a-p                      ! GIVE BACK THE REFLECTION POINTS COORDINATE

R(2)=2*b-q

END SUBROUTINE REFLECTION

Answer 2

Reflection of point A(x,y) in the line y=mx+c.
Given point P(x,y) and a line L1 y=mx+c.
Then P(X,Y) is the reflected point on the line L1.
If we join point P to P’ to get L2 then gradient of L2=-1/m1 where m1 is gradient of L1.

  L1 and L2 are perpendicular to each other.
        therefore,
    Get the point of intersection of L1 and L2 say m(a,b)
    Since m(a,b) is the midpoint of PP’ i.e. L2, then
    M= (A+A')/2 
    i.e. m(a,b)=(A(x,y)+ A^' (x^',y^' ))/2.
      from this we can get coordinates of  A^' (x^',y^' )

Example

Find the image of point P(4,3) under a reflection in the line  y=x-5
M1=1
M2 will be -1
Equ. L2 with points, (4,3) , (x ,y) grad -1 is
     y=-x+7
Point of intersection say, M(a ,b)
 Note that, L1 =L2 ;
  Then x-5=-x+7
  This gives the point for M that is M( 6,1)
      Then;
         M(6,1)=(P(4,3)+P^' (x^',y^' ))/2
          M(6,1)=[(4+x)/2  ,(3+y)/2]
              This gives x = 8 and y = -1 hence,
                  P^' (x^',y^' )=         P^' (8,-1)

Answer 3

Ok, I'm going to give you a cookbook method to do this. If you're interested in how I derived it, see http://www.sdmath.com/math/geometry/reflection_across_line.html#formulasmb

Given point (x1, y1) and a line that passes through (x2,y2) and (x3,y3), we can first define the line as y = mx + c, where:

slope m is (y3-y2)/(x3-x2)

y-intercept c is (x3*y2-x2*y3)/(x3-x2)

If we want the point (x1,y1) reflected through that line, as (x4, y4), then:

set d = (x1 + (y1 - c)*m)/(1 + m^2) and then:

x4 = 2*d - x1
y4 = 2*d*m - y1 + 2*c