New modules - euclidean-intersection & euclidean-distance

[DenisCarriere]

New modules - euclidean-intersection & euclidean-distance

Found in @turf/point-to-line-distance, it would be nice to extract those modules out as their own seperate modules.

Euclidean distance does not account for the curvature of the earth which would be known as a straight line distance.

Module name proposal

• @turf/euclidean-intersection
• @turf/euclidean-distance

or all methods grouped under @turf/euclidean

Current code

/**
 * Returns euclidean distance between two points
 *
 * @private
 * @param {Object} from first point
 * @param {Object} to second point
 * @param {Object} [options={}] Optional parameters
 * @param {string} [options.units='kilometers'] can be degrees, radians, miles, or kilometers
 * @returns {number} squared distance
 */
function euclideanDistance(from, to, options) {
    var units = options.units;
    // translate points if any is crossing the 180th meridian
    var delta = 0;
    if (Math.abs(from[0]) >= 180) {
        delta = (from[0] > 0) ? -180 : 180;
    }
    if (Math.abs(to[0]) >= 180) {
        delta = (to[0] > 0) ? -180 : 180;
    }
    var p1 = toMercator([from[0] + delta, from[1]]);
    var p2 = toMercator([to[0] + delta, to[1]]);

    var sqr = function (n) { return n * n; };
    var squareD = sqr(p1[0] - p2[0]) + sqr(p1[1] - p2[1]);
    var d = Math.sqrt(squareD);
    return convertDistance(d, 'meters', units);
}

/**
 * Returns the point H projection of a point P on a segment AB, on the euclidean plain
 * from https://stackoverflow.com/questions/10301001/perpendicular-on-a-line-segment-from-a-given-point#answer-12499474
 *            P
 *           |
 *           |
 *  _________|____
 * A          H   B
 *
 * @private
 * @param {Array<number>} a first segment point
 * @param {Array<number>} b second segment point
 * @param {Array<number>} p external point
 * @returns {Array<number>} projected point
 */
function euclideanIntersection(a, b, p) {
    var x1 = a[0], y1 = a[1],
        x2 = b[0], y2 = b[1],
        x3 = p[0], y3 = p[1];
    var px = x2 - x1, py = y2 - y1;
    var dab = px * px + py * py;
    var u = ((x3 - x1) * px + (y3 - y1) * py) / dab;
    var x = x1 + u * px, y = y1 + u * py;
    return [x, y]; // H
}