距離をずらした後の座標（x1、y1）、（x2、y2）を持つ線の新しい点（x3、y3）、（x4、y4）の計算方法は？

Question

calculate new points(x3,y3),(x4,y4) of the line(x1,y1),(x2,y2) after shifting d distance and it should always parallel

Answer 1

このようなものはありますか？

function lineOffset(x1, y1, x2, y2, d){ 
    // delta between points 
    var dx = x2 - x1; 
    var dy = y2 - y1; 

    // gradient of line segment 
    var dx_dy = dx/dy; 
    // inverse (perpendiular line) 
    var dy_dx = 1/dx_dy; 

    // point inbetween P1 and P2 
    var x12 = (x1 + x2)/2; 
    var y12 = (y1 + y2)/2; 

    // point inbetween P3 and P4 
    var x34 = x12 + (d * Math.cos(dy_dx)); 
    var y34 = y12 + (d * Math.sin(dy_dx)); 

    // sub 1/2 the line length to get P3 
    var x3 = x34 - (dx/2); 
    var y3 = y34 - (dy/2); 

    // add 1/2 the line length to get P4 
    var x4 = x34 + (dx/2); 
    var y4 = y34 + (dy/2); 

    return { 
     x3: x3, 
     y3: y3, 
     x4: x4, 
     y4: y4 
    } 
}