algorithm to find a certain lat long position in a polygon enclosed by lat/long postions or not

Question

I have a set of polygons drawn on google map,now I want to implement the algo that given a postion with certain lat/long in which of the polygon it lies.

Note:polygons are also drawn given the lat/long positions using the google maps api

So Is there any api for it or how can i convert lat/long positions to x-y planes so that i can check if a given point lies in which area using area formulas?

Answer 1

would google.maps.geometry.poly.containsLocation work ?

Answer 2

use below ray casting alogrithm that may help u to resolve the problem

google.maps.Polygon.prototype.Contains = function(point) {
        // ray casting alogrithm
        var crossings = 0,
            path = this.getPath();
        // for each edge
        for (var i = 0; i < path.getLength(); i++) {
            var a = path.getAt(i),
                j = i + 1;
            if (j >= path.getLength()) {
                j = 0;
            }
            var b = path.getAt(j);
            if (rayCrossesSegment(point, a, b)) {
                crossings++;
            }
        }

        // odd number of crossings?
        return (crossings % 2 == 1);

        function rayCrossesSegment(point, a, b) {
            var px = point.lng(),
                py = point.lat(),
                ax = a.lng(),
                ay = a.lat(),
                bx = b.lng(),
                by = b.lat();
            if (ay > by) {
                ax = b.lng();
                ay = b.lat();
                bx = a.lng();
                by = a.lat();
            }
            // alter longitude to cater for 180 degree crossings
            if (px < 0) { px += 360 };
            if (ax < 0) { ax += 360 };
            if (bx < 0) { bx += 360 };

            if (py == ay || py == by) py += 0.00000001;
            if ((py > by || py < ay) || (px > Math.max(ax, bx))) return false;
            if (px < Math.min(ax, bx)) return true;

            var red = (ax != bx) ? ((by - ay) / (bx - ax)) : Infinity;
            var blue = (ax != px) ? ((py - ay) / (px - ax)) : Infinity;
            return (blue >= red);
        }
    };