Find a point on perpendicular of a line - math

Question

I have a side of a irregular polygon (x1,y1) (x2,y2) at a angle A and the midpoint of the side (mx, my).
I need to find two points (x3,y3) and (x4,y4) on a perpendicular from A passing through (mx, my), with some offset. So that I could check which point is the inside/outside of the polygon.
I'll use the outside one to show the measurement text of the side of the polygon, e.g 2cms
Click to see Visuals

Answer 1

You dont specify the language, but in almost all of them we have atan2 function for that.
canvas.width = window.innerWidth - 10;
canvas.height = window.innerHeight - 10;
const ctx = canvas.getContext('2d');
const line = [canvas.width/2, canvas.height/2, 10, 10];
const pLen = 100; // Length of perpendicular
function drawOrto(line) {
// line vector
const dx = line[2] - line[0];
const dy = line[3] - line[1];
// center point
const mx = line[0] + dx / 2;
const my = line[1] + dy / 2;
const atan = Math.atan2(dy, dx);
// perpendicular vector
const pdx = - pLen * Math.sin(atan);
const pdy = pLen * Math.cos(atan);
ctx.clearRect(0, 0, canvas.width, canvas.height);
// Line
ctx.beginPath();
ctx.strokeStyle = 'black';
ctx.moveTo(line[0], line[1]);
ctx.lineTo(line[2], line[3]);
ctx.stroke();
// Perpendicular (draw vector both sides from center)
ctx.beginPath();
ctx.strokeStyle = 'red';
ctx.moveTo(mx - pdx / 2, my - pdy / 2);
ctx.lineTo(mx + pdx / 2, my + pdy / 2);
ctx.stroke();
// Dots
ctx.fillStyle = 'red';
ctx.fillRect(line[0] - 1, line[1] - 1, 3, 3);
ctx.fillRect(line[2] - 1, line[3] - 1, 3, 3);
ctx.fillStyle = 'yellow';
ctx.fillRect(mx - 1, my - 1, 3, 3);
ctx.fillStyle = 'green';
ctx.fillRect(mx - pdx / 2 - 1, my - pdy / 2 - 1, 3, 3);
ctx.fillRect(mx + pdx / 2 - 1, my + pdy / 2 - 1, 3, 3);
}
drawOrto(line);
canvas.onmousemove = e => {
line[2] = e.offsetX;
line[3] = e.offsetY;
drawOrto(line);
}
<canvas id=canvas></canvas>

Answer 2

Rotation by a quarter-turn is easy: if you have a vector with coordinates (x, y) and rotate it by a quarter-turn anti-clockwise, the coordinates of the new vector are (-y, x).
To find a point (x5,y5) on the perpendicular, you can rotate point (x2,y2) by a quarter-turn anti-clockwise centered on (mx,my), so that:
x5 - mx = - (y2 - my)
y5 - my = x2 - mx
Then you can choose your two points (x3, y3) and (x4, y4) by renormalizing vector (x5-mx, y5-my) to have the length you want:
x4 - mx = (x5 - mx) * (2 cm) / ((x5-mx)**2 + (y5-my)**2)
y4 - my = (y5 - my) * (2 cm) / ((x5-mx)**2 + (y5-my)**2)
x3 - mx = (x5 - mx) * (-2 cm) / ((x5-mx)**2 + (y5-my)**2)
y3 - my = (y5 - my) * (-2 cm) / ((x5-mx)**2 + (y5-my)**2)

Answer 3

First you have to find the side equation:
y - y1 = (y2-y1)/(x2-x1)*(x-x1)
y = (y2-y1)/(x2-x1) * x1 - (y2-y1)/(x2-x1) * x + y1
Once you have an equation like this y=mx+q (y and x are costants, so you only care about finding m and q) you can find the middle point
M((x1+x2)/2), (y1+y2)/2)
When you have both the equation and the middle point you can choose your offset. Now just find the points that correspond to that offset on the segment
P1(mx + offset, my + (offset * m)) and P2(mx - offset, my - (offset * m))
where m is the one you found in the first equation
Now you have the 2 points
To find out which is inside and which is outside we would need to have more data. You could try for example to bisect the angles to find the middle point of the polygon M1 and check which between P1 and P2 has a distance to M1 so that:
if distance(P1, M1) < distance(M, M1) then P1 is inside
"if the distance between the point and the middle of the polygon is less than the distance between the middle of the polygon and the middle of the side then P1 is inside the polygon"
There are other solutions, but with the data available this is all i can do. Now you just have to turn this explanation into code
LEGEND:
y -> equation constant
x -> equation constant
m -> line inclination
q -> intersection with axis (you really don't need it for this)
M -> middle point of the side mx, my -> coordinates of middle point of the side
P1 -> point on the line with positive offset
P2 -> point on the line with negative offset
M1 -> middle point of the polygon