java HttpExchange create a URI redirect URL link

Question

I am trying to create a URI redirect link. But, I simply cannot get the syntax right. I assume it's really easy, and that's why I can't find similarly asked questions. Either that, or it's not a typical user case. In either event, here is my method:

    public void handle (final HttpExchange exchange) throws IOException {
        try {
            final InetSocketAddress redirectServerAddress = this.selectRedirectServerAddress(exchange.getRemoteAddress().getAddress());

            final URI requestURI = exchange.getRequestURI();

            // TODO: create the redirect URI
            // I thought this would be most promising: 
            // URI redirectURI = new URI(http:redirectServerAddress, "/path/here", null)
            URI redirectURI = null;
                
            // send the exchanges's response headers using the code 307 (temp redirect) and zero as response length
            exchange.getResponseHeaders().set("Location", redirectURI.toASCIIString());
            exchange.sendResponseHeaders(307, 0);
            
            Logger.getGlobal().log(Level.INFO, "Redirected request for \"{0}\" to \"{1}\".", new URI[] { requestURI, redirectURI });
        } finally {
            exchange.close();
        }
    }

I have tried various options, the one answer here looks promising, but I could not get it to work. Any help would be appreciated. Thanks!