The points P(2,-4) and Q(6,0) are consecutive vertices of the parallelogram which has its centre at the origin. Find:
a) The other two vertices
b) The angles of the parallelogram  

1.- Change the coordinates of the vertices R and S, in this way the quadrilateral PQRS will be a parallelogram with its centre at the origin O. In this way you already obtain the coordinates of the vertices R and S 

2.- Calculate the cosine of the angle formed by the vectors RP and RS, and subsequently the said angle  

3.- Verify the result in the figure

Remember that the diagonals of a rhombus intersect at right angles at their mid point.
Also you should remember that the area of a rhombus is: 

Steps to follow:
1.- Find the equation of the line AC, of which we know the point M (mid point of BD) and the direction vector AC (perpendicular to the vector BD)
2.- The point A, belongs to the line AC and we know it can be found on the x-axis.
3.- Find the coordinates of the point C, knowing that the vectors AM and MC are equal.

Clearly the quadrilateral ABCD shown in this figure is not a rhombus. 

You can change the x coordinate of the point A, and the coordinates of the point C, to achieve what you want. 

If you have already solved the problem, you can check your solutions in the figure. 

Note: The area which appears is not correct until effectively the quadrilateral becomes a rhombus .

In this figure you can change the coordinates of the point A, as the problem asks.
To be the bisector of the segment OA, the given line r: 2x+y-4=0 cuts it at the mid point, M.

But this is not sufficient, furthermore r has to be perpendicular to OA. Therefore the initial figure does not give us the coordinates of the required end A of the segment. 

The equation of the line perpendicular to r, which passes through O, must be found. That is where we will find A. 

Then find the point of intersection of that line with r, which will be M. 

Remember that a trapezium is a quadrilateral with two sides parallel and the other two are not. And if it is isosceles the two sides which are not parallel, are equal.
The area of a trapezium is, where B=longest base (longest parallel side), b=shortest base (shortest parallel side) and h=height (length of the segment perpendicular to the two bases)

In this figure the given line, x+y-2=0 is drawn, and one line which passes through the point P(0,5). But it isn't parallel to the one given, that is the quadrilateral drawn is not a trapezium.
1.- Firstly you have to find the equation of the line parallel to x+y-2=0, and that passes through P(0,5) 

2.- Knowing the gradient, m, of this line, and that it passes through P(0,5), you can find its equation. 

3.- The intersection of the two lines with the axes permits you to know the coordinates of the four vertices of the trapezium. 

4.-You can introduce the value of m in the figure. In this form the quadrilateral is now a trapezium , and furthermore isosceles. And now you can see if you have calculated the equation of the line and the coordinates of the vertices correctly. You can see that in the initial figure, the angles formed by the segment AB with the lines (bases), were not two of 90º. But by making the lines parallel, now they are. 

5.- Now the only thing left to do is find the lengths of the two bases (distance between two points), and the height (distance from a point to a line), to find the area. You can verify its value in the figure.

Note: The value of the area which appears in the figure does not correspond with that of the initial quadrilateral, although it will be the value which appears when it becomes a trapezium.