1. Jul 30, 2016

### gracy

While proving the Midpoints of the Sides of a Quadrilateral Form a Parallelogram , I got bogged down with position vectors.

Let a,b,c and d be the position vectors of A,B,C and D. But where is the origin? Aren't we supposed to locate position of origin?

2. Jul 30, 2016

### PeroK

You can put the origin wherever you like. I might put it at point $A$.

3. Jul 30, 2016

### gracy

If we take origin at A, position vector of A that is given to be a will be 0,0 . Right?

4. Jul 30, 2016

### cnh1995

As PeroK said, you can put the origin at any point as per your convenience.
This problem can be solved using simple properties of triangle.

5. Jul 30, 2016

### PeroK

I'd say the position vector of $A$ in that case is $\vec{0}$. This may simplify the problem.

6. Jul 30, 2016

### gracy

I want to use the following formula for position vector of mid point

For that I need origin other than point A.

7. Jul 30, 2016

### PeroK

That's the right formula, but it's even simpler with $\vec{OA} = \vec{0}$.

8. Jul 30, 2016

### gracy

If we take $\vec{OA}$ = $\vec{0}$
The formula will be reduced to
$\vec{OM}$ = $\frac{OB}{2}$
(I meant position vector of OB , I don't know how to get vector sign on top of OB)

9. Jul 30, 2016

### PeroK

Okay, that gives you the position vector of point $P$.

Have you thought yet about what you need to do to show that $PQRS$ is a parallelogram?

10. Jul 30, 2016

### gracy

In the book it's given
PQ= position vector of Q - position vector of P
How so? Is there any particular standard formula for this that I am missing?

11. Jul 30, 2016

### PeroK

It's not a formula. But, what defines a parallelogram?

12. Jul 30, 2016

### gracy

A Parallelogram has opposite sides parallel and equal in length.

13. Jul 30, 2016

### PeroK

Good. Think a bit more about what you need to do to show this.

You can get from the origin to point $Q$ in two ways:

$\vec{OQ}$

Or:

$\vec{OP} + \vec{PQ}$

Therefore:

$\vec{OQ} = \vec{OP} + \vec{PQ}$

14. Jul 30, 2016