如何使用 html 画布和打字稿绘制给定两个点的三角形

Question

I am sure I have missed something obvious, but I am trying to draw a quadratic curve between two points using an html canvas, for which I need a 'control point' to set the curve. The start and end point of the curve are known, the control point is unknown because the lines are dynamically rotated. I just need to find this third point of the triangle in order to set the control point

I use this function to find the mid point of the line:

lineMidPoint(p: Point, q: Point): Point {

    let x = (p.x + q.x) / 2;

    let y = (p.y + q.y) / 2; 

    return { x: x, y: y } as Point;
}

This function works as expected.

Then a second function to get the angle of the line relative to the origin:

getAngleRelativeToOrigin(start: Point, end: Point): number {

    let dx = start.x - end.x;

    let dy = start.y - end.y;

    let radians = Math.atan2(dy, dx);

    return radians * (180/Math.PI);
}

It is hard to verify that this function is working.

Then finally I have a function for rotating the midpoint around either the start or the end of the line in order to find the control point:

getControlPoint(start: Point, end: Point): Point {

    let midPoint = this.lineMidPoint(start, end);

    let offset = 45 * (Math.PI / 180);

    let theta = this.getAngleRelativeToOrigin(start, end) + offset;

    let x = Math.cos(theta) * (start.x - midPoint.x) - Math.sin(theta) * (start.y - midPoint.y) + midPoint.x;

    let y = Math.sin(theta) * (start.x - midPoint.x) - Math.cos(theta) * (start.y - midPoint.y) + midPoint.y;

    return { x: x, y: y } as Point;
}

The result is this:

Those lines that are not connected to circles (for instance on the far right) should all be the length of the line they start from / 2, but they are clearly inconsistent.

When I draw the quadratic curves they are all wonky:

Can anyone lend a hand and tell me where Ive gone wrong?

Answer 1

OK, your middle point is correct. Now determine difference vector and perpendicular to the line

let dx = start.x - end.x;
let dy = start.y - end.y;
let leng = Math.hypot(dx, dy);
let px = - dy / leng;   //and get perpendicular unit vector
let py = dx / leng;

I am not sure what logic you wanted to implement, so I propose to get control point at distance d from line middle (so curve is symmetrical)

let xxx = midPoint.x + d * px;
let yyy = midPoint.y + d * py;

If you want to rotate middle point about start point, it might be done using the next approach:

let cost = Math.cos(45 * (Math.PI / 180));
let sint = Math.sin(45 * (Math.PI / 180));

let x = start.x + 0.5 * dx * cost - 0.5 * dy * sint;
let y = start.y + 0.5 * dx * sint + 0.5 * dy * cost;