How do you find a point at a given perpendicular distance from a line?

Question

I have a line that I draw in a window and I let the user drag it around. So, my line is defined by two points: (x1,y1) and (x2,y2). But now I would like to draw "caps" at the end of my line, that is, short perpendicular lines at each of my end points. The caps should be N pixels in length.

Thus, to draw my "cap" line at end point (x1,y1), I need to find two points that form a perpendicular line and where each of its points are N/2 pixels away from the point (x1,y1).

So how do you calculate a point (x3,y3) given it needs to be at a perpendicular distance N/2 away from the end point (x1,y1) of a known line, i.e. the line defined by (x1,y1) and (x2,y2)?

Answer 1

You need to compute a unit vector that's perpendicular to the line segment. Avoid computing the slope because that can lead to divide by zero errors.

dx = x1-x2 dy = y1-y2 dist = sqrt(dx*dx + dy*dy) dx /= dist dy /= dist x3 = x1 + (N/2)*dy y3 = y1 - (N/2)*dx x4 = x1 - (N/2)*dy y4 = y1 + (N/2)*dx 

Answer 2

You just evaluate the orthogonal versor and multiply by N/2

vx = x2-x1 vy = y2-y1 len = sqrt( vx*vx + vy*vy ) ux = -vy/len uy = vx/len  x3 = x1 + N/2 * ux Y3 = y1 + N/2 * uy  x4 = x1 - N/2 * ux Y4 = y1 - N/2 * uy