Get virtualenv's bin folder path from script

I'm using virtualenvwrapper with a django project that has a management task that automatically writes some config files, so the user just has to ``` ./manage.py generate_configuration > much_nice.conf ``` And then move the file elsewhere. One of the generated config files is a task for supervisord that launches a celery worker. The problem I'm getting is that I don't know how to output the path of the celery executable that is within the bin folder of the virtualenv. Essentially, I'd like to have the output of the command ``` which celery ``` One option is using `sys.executable`, get the folder (which seems to be the `bin` folder of the virtualenv) and that's it... but I'm not sure. Doesn't virtualenv have any kind of method to get the path itself?