Transfer coordinates to a new system

Question

I would like to convert a point coordinates to a new generated coordinate system the original system start in the top left corner of the image (0,0)

The information that I have in a new system are : 1- I have the value of the new original (x0,y0) in some where in the image 2- also I have 2 points on both new axes ( 4 points in total 2 in each line) using this I can calculate the line equation for the 2 lines of axes (y=a1x+b1) ,(y=a2x+b2) 3- I have vector for each line (Vx, Vy)

Note: sometime the new axes rotate (the lines are not exactly horizontal or vertical)

How can I convert the points coordinates to this new system

any help will be so appreciated

here is the image

Answer 1

First express your lines like a1*(x-x0)+b1*(y-y0)=0 and a2*(x-x0)+b2*(y-y0)=0 and their intersection x0,y0 is accounted for already in the equations.

updated signs

The transformation from x,y to z,w is

z = -sqrt(a1^2+b1^2)*(a2*(x-x0)+b2*(y-y0))/(a2*b1-a1*b2)
w =  sqrt(a2^2+b2^2)*(a1*(x-x0)+b1*(y-y0))/(a1*b2-a2*b1)

and the inverse

x = x0 - b1*z/sqrt(a1^2+b1^2) + b2*w/sqrt(a2^2+b2^2)
y = y0 + a1*z/sqrt(a1^2+b1^2) - a2*w/sqrt(a2^2+b2^2)

It would be helpful to scale the coefficients such that sqrt(a1^2+b1^2)=1 and sqrt(a2^2+b2^2)=1.

Note that this works for non-orthogonal lines also. As long as they are not parallel and a2*b1-a1*b2!=0 it is going to work.

Example

The z line (-2)*(x-3) + (1)*(y-1) = 0 and w line (-1)*(x-3) + (-4)*(y-1) = 0 meet at (3,1). The coefficients are thus a1=-2, b1=1, a2=-1, b2=-4.

The coordinates (x,y)=(2,1) transform to

z = -sqrt((-2)^2+1^2) ((-1) (x-3)+(-4) (y-1))/((-1) 1-(-2) (-4)) = 0.2484
w = sqrt((-1)^2+(-4)^2) ((-2) (x-3)+1 (y-1))/((-2) (-4)-(-1) 1) = 0.9162

With the inverse

x = -1 z/sqrt((-2)^2+1^2)+(-4) w/sqrt((-1)^2+(-4)^2)+3 = 2
y = (-2) z/sqrt((-2)^2+1^2)-(-1) w/sqrt((-1)^2+(-4)^2)+1 = 1

Development

For a line a1*(x-x0)+b1*(y-y0)=0 the direction vector along the line is e1 = [e1x,e1y]= [-b1/sqrt(a1^2+b1^2),a1/sqrt(a1^2+b1^2)]. Similarly for the other line.

The screen coordinates of a local point [z,w] is found by starting at the origin x0, y0 and moving by z along the first line and then by w along the second line. So

x = x0 + e1x*z + e2x*w = x0 -b1/sqrt(a1^2+b1^2)*z - b2/sqrt(a2^2+b2^2)*w
y = y0 + e1y*z + e2y*w = y0 +a1/sqrt(a1^2+b1^2)*z + a2/sqrt(a2^2+b2^2)*w

Now I need to flip the direction of the second line to make it work per the original posting visualization, by reversing the sign of w.

To find z, w from x, y invert the above two equations.