I want to read only one line from the code so by using node.js I:

import fs = require('fs');

then:

var output = fs.readFileSync('/direction/projectname/server.js','utf8');
console.log(output);

I output this way whole file from this direction... BUT I WANT TO PRINT ONLY ONE SPECIFIC LINE OF THAT SOURCE CODE, for example, 51st line.

Any idea how to do it?