通过点的平行线

Question

我有一个Line(P1, P2)和一个点NP(x,y)，我想要的是通过Line(A,B)创建平行的NP 我需要JavaScript中的一个函数来获取Line（A，B）

已解决：我在@MBo的帮助下用JavaScript代码编写了函数

function parallel(pt1, pt2, newPoint) {
	let NP = new THREE.Vector2(newPoint.x, newPoint.y)
	let p1 = new THREE.Vector2(pt1.x, pt1.y)
	let p2 = new THREE.Vector2(pt2.x, pt2.y)
	let dir = new THREE.Vector2(p2.x-p1.x, p2.y-p1.y);
	let W = new THREE.Vector2(NP.x-p1.x, NP.y-p1.y);
	let dw = dir.clone().dot(W);
	let dd = dir.clone().dot(dir);
	let dmw = dir.clone().multiplyScalar(dw).divideScalar(dd);
	let PP = p1.clone().add(dmw)
	let V = NP.clone().sub(PP)
	let pi1 = p1.add(V)
	let pi2 = p2.add(V)
	return {A:{x:pi1.x,y:pi1.y}, B:{x:pi2.x,y:pi2.y}}
}

Answer 1

找到 NP 的投影到 P1P2 线上。

首先获得向量

Dir = (P2.X-P1.X, P2.Y-P1.Y)
W =   (NP.X-P1.X, NP.Y-P1.Y)

使用标量（点）预测法将 NP 投影到 P1P2 的点是：

PP = P1 + Dir * (Dir.dot.W) / (Dir.dot.Dir)

和投影向量

V = NP - PP

现在并行段的末端是

P1' = P1 + V
P2' = P2 + V

Answer 2

先让P1=[p1x,p1y]，P2=[p2x,p2y]，NP=[nx,ny]然后

W = [wx,wy] = (P2-P1)/2 = [(p2x-p1x)/2, (p2y-p1y)/2]

和

A = NP-W = [nx-wx, ny-wy]
B = NP+W = [nx+wx, ny+wy]

JS代码

function parallel([p1x,p1y],[p2x,p2y],[nx,ny]) {
  let wx = (p2x-p1x)/2;
  let wy = (p2y-p1y)/2;
  return [[nx-wx,ny-wy],[nx+wx,ny+wy]]
}


// TEST

let P1 = [2,3]; 
let P2 = [6,5];
let NP = [5,2];
let [A,B] = parallel(P1,P2,NP);

// print
console.log(`for P1=[${P1}], P2=[${P2}], NP=[${NP}] 
result is A=[${A}], B=[${B}]`);