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

First the perpendicular line has slope that is the negative reciprocal.

so the cap has slope (x1-x2)/(y2-y1) , which is the negative reciprocal of the slope of given your line.

We now have four unknowns: x3, y3, x4, y4 where (x3,y3) and (x4,y4) are the endpoints of the cap.

So we need 4 equations.

Midpoint formula gives:

(x3 + x4)/2 = x1 ----> x3 + x4 = 2 * x1

(y3 + y4)/2 = y1 ---> y3 + y4 = 2 * y1

Distance formula gives:

( x3 - x4) ^2 + ( y3-y4)^2 = N^2

Finally the slopes are negative reciprocals of each other, so

(y4 - y3)/(x4-x3) = (x1-x2)/(y2-y1) per the formula above.

Then y4 - y3 = (x1-x2)*(x4-x3)/(y2-y1) <--- please label this equation ALPHA

Plugging this into the distance formula:

(x3-x4)^2 [ 1 + {(x1-x2)/(y2-y1)}^2 ] = N^2

So then

(x3-x4) = N/ sqrt[ 1 + {(x1-x2)/(y2-y1)}^2 ]

Adding this to the second equation gives:

2*x3 = 2*x2 + N/ sqrt[ 1 + {(x1-x2)/(y2-y1)}^2 ]

so that

x3 = x2 + (1/2) N/ sqrt[ 1 + {(x1-x2)/(y2-y1)}^2 ]

Once found, the second equation gives x4

You can then plug the values you have just calculated for x3 and x4 into the slope

equation labeled ALPHA above. That will give you a numerical value for y4 - y3

Finally, using y3 + y4 = 2*y2, you can find y3 and y4