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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def equirectangular_distance(self, other):
 """ Return the approximate equirectangular when the location is close to the center of the cluster. For small distances, Pythagoras’ theorem can be used on an equirectangular projection. Equirectangular formula:: x = Δλ ⋅ cos φm y = Δφ d = R ⋅ √(x² + y)² It will always over-estimate compared to the real Haversine distance. For example it will add no more than 0.05382 % to the real distance if the delta latitude or longitude between your two points does not exceed 4 decimal degrees. The standard formula (Haversine) is the exact one (that is, it works for any couple of longitude/latitude on earth) but is much slower as it needs 7 trigonometric and 2 square roots. If your couple of points are not too far apart, and absolute precision is not paramount, you can use this approximate version (Equirectangular), which is much faster as it uses only one trigonometric and one square root:: Python 2.7.6rc1 (v2.7.6rc1:4913d0e9be30+, Oct 27 2013, 20:52:11) [GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin Type "help", "copyright", "credits" or "license" for more information. 5.62202811241 529.424701041 2.78262710571 529.424701073 @param other: a ``GeoPoint`` instance. @return: the great-circle distance, in meters, between this geographic coordinates to the specified other point. """

        x = math.radians(other.longitude - self.longitude) \
                * math.cos(math.radians(other.latitude + self.latitude) / 2);
        y = math.radians(other.latitude - self.latitude);
        return math.sqrt(x * x + y * y) * GeoPoint.EARTH_RADIUS_METERS;