get coordinates of arrow of a line point - vector

Question

Consider I have 2 points.
QPoint(100,100)
QPoint(200,200)
Now, I need to draw a pointed arrow at the end of line QPoint(200,200).
How can I get the points of the arrow coordinates since the line is in inclined angle? The arrow should be like given below.
As this is more a general question, Im tagging on pyqt5 and css.

Answer 1

You need to calculate the slope of your line, which will let you find a point on the line a given distance from your endpoint. You can then build a a new line perpendicular to the original line that goes through the point. The ends of the arrows should lie on that line a given distance from the original. It's easier to show than explain:
function draw(point1, point2, distance, length) {
// slope is dx/dy
let dx = point2[0] - point1[0]
let dy = point2[1] - point1[1]
let v_norm = Math.sqrt(dx ** 2 + dy ** 2)
// point on line at distance
let point_on_line = [point2[0] - distance * dx / v_norm, point2[1] - distance * dy / v_norm]
// endpoints of arrows at length above point (the distance from the original line
let point_below = [point_on_line[0] - length * -dy / v_norm, point_on_line[1] - length * dx / v_norm, ]
let point_above = [point_on_line[0] + length * -dy / v_norm, point_on_line[1] + length * dx / v_norm, ]
var canvas = document.getElementById('canvas');
var ctx = canvas.getContext('2d');
ctx.beginPath();
ctx.moveTo(...point1);
ctx.lineTo(...point2);
ctx.moveTo(...point_above)
ctx.lineTo(...point2)
ctx.lineTo(...point_below)
ctx.stroke();
}
draw([100, 100], [200, 150], 20, 10)
draw([100, 100], [300, 150], 20, 10)
draw([100, 100], [150, 10], 20, 10)
draw([100, 100], [90, 150], 20, 10)
draw([100, 100], [200, 100], 20, 10)
draw([100, 100], [5, 10], 20, 10)
<canvas id="canvas" width='500' height='500'></canvas>
You can alter the shape of the arrows using distance and length

Answer 2

If we have a look at the points we need (U and D needed, S and E given):
U
/|
/ |
S---------------------E
\ |
\|
D
We actually have to go back from S to E, then we turn by 90 degrees and walk up the same length. If we take the direction between S and E (represented as Vectors):
d = E - S
Then we can get the distance to walk back:
b = -d / 4
And then to walk up in a 90 degree angle, we just swap x and y, to go down we just take the negative one:
u = [b.y, b.x]
d = -u
So we can finally get our points as:
U = S + b + u
D = S + b + d
(The code above is pseudocode as you wanted a general solution)