JS判断点是否在线段上的代码

/**
 * 判断点c是否在ab组成的线段上
 * @param {x,y} a 点
 * @param {x,y} b 点
 * @param {x,y} c 点
 * @returns boolean
 */
function isPointOnLineSegment(a, b, c) {
  // 计算向量ab和ac的叉积
  let crossProduct = (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
  // 如果不等于0，说明不共线，直接返回false
  if (crossProduct !== 0) {
    return false;
  }
  // 否则，检查c点是否在ab线段的范围内
  return (
    Math.min(a.x, b.x) <= c.x &&
    c.x <= Math.max(a.x, b.x) &&
    Math.min(a.y, b.y) <= c.y &&
    c.y <= Math.max(a.y, b.y)
  );
}

// 测试
const a = {x:0,y:0}
const b = {x:0,y:1}
const c = {x:0,y:2}
const d = {x:1,y:1}
const e = {x:1,y:2}
const f = {x:2,y:2}

console.log(isPointOnLineSegment(a, c, b)); // true
console.log(isPointOnLineSegment(a, f, d)); // true
console.log(isPointOnLineSegment(c, f, e)); // true
console.log(isPointOnLineSegment(a, b, c)); // false
console.log(isPointOnLineSegment(a, f, c)); // false
console.log(isPointOnLineSegment(a, c, f)); // false