在线查找点的新坐标

Question

假设我知道两点:x1,y1和x2,y2.我知道我可以很容易地用毕达哥拉斯计算这条线的长度,但如果我想计算一条缩短版本的线怎么办呢.例如,我希望x,y坐标朝向下一个点10个像素.有没有一个简单的公式来找到一个有角度的线上的任何新点？

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好的,这是作为JavaScript函数的解决方案.为了解释的目的,我故意将这个额外的冗长.如下面的评论所示,您必须首先找到角度,然后用新的斜边计算边.

/**
 *  Calculates length of a line on Cartesian grid, then returns point that is X number of pixels down the line.
 *
 * @param  {Number} fromX - starting x point
 * @param  {Number} fromY - starting y point
 * @param  {Number} toX - ending x point for vector
 * @param  {Number} toY - ending y point for vector
 * @param  {Number} pxDistance - Number of pixels down line toward ending point to return
 * @return {Object} Returns x/y coords of point on line based on number of pixels given
 */ 
function stortenLineDistance(fromX, fromY, toX, toY, pxDistance){

    //if line is vertical
    if(fromX === toX)
        return {x: toX, y: toY > fromY ? fromY + pxDistance : fromY - pxDistance};

    //if line is horizontal
    if(fromY === toY)
        return {y: toY, x: toX > fromX ? fromX + pxDistance : fromX - pxDistance};

    //get each side of original triangle length
    var adjacent   = toY - fromY;
    var opposite   = toX - fromX;
    var hypotenuse = Math.sqrt(Math.pow(opposite, 2) + Math.pow(adjacent,2));

    //find the angle
    var angle = Math.acos(adjacent/hypotenuse);

    //calculate new opposite and adjacent sides
    var newOpposite = Math.sin(angle) * pxDistance;
    var newAdjacent = Math.cos(angle) * pxDistance;

    //calculate new x/y, see which direction it's going
    var y = fromY - newAdjacent,
        x = fromX + newOpposite;

    return {y: y, x: x};

}

编辑:哇,Stackoverflow是残酷的.如果你愿意,可以随意删除这个问题,不过我觉得这对那些坚持同一问题的人有帮助.也许我错了.

无论如何,感谢评论者的帮助.

Answer 1

我创造了一个利用相应边的比例的小提琴.将变量调整smallerLen到任何单位以查看点在线上移动.

http://jsfiddle.net/3SY8v/