Let P, Q, R, and S be the points on the plane with position
vectors negative 2i minus j, 4i, 3i plus 3j, and negative 3i
plus 2j, respectively.
The quadrilateral PQRS must be a--
So let's just graph each of these positions vectors,
or graph the points that they're specifying.
So let me draw my coordinate axes right over here.
So this is my vertical.
I can draw a straighter line that.
There you go.
That's about as good as I can do.
Let me draw the horizontal.
That's a better job, I think.
And then let me mark it off.
Let's see.
We go as high as 4.
We go up to 4i.
So 1, 2, 3, 4.
We goes as low was negative 2i.
So 1, 2.
Just making sure I make it a little bit
even, because we're going to have to figure out
the shape of whatever we draw.
Then we're going to go up 3.
1, 2, 3.
And we actually go down 1, as well.
So I think that'll give us-- and we could keep marking them off
if we wanted to, but we don't need to,
based on the positions that have been specified.
So let's draw these vectors.
So let's draw vector P first.
Negative 2i minus j.
So negative 2i, i is along the horizontal,
or as you can imagine, it's the x-coordinate.
So negative 2i minus j.
So it is this position vector right over here.
If we put it in standard form, starting
with its base at the origin.
It's that position vector.
But it's specifying that point over there,
which is the coordinate negative 2, negative 1.
Let me just write it down.
Negative 2, negative 1.
Now let's do 4i.
So 4i is literally-- that's just 4 in the x direction.
1, 2, 3, 4.
There's no j component, no vertical component.
So it specifies that position right over there,
or it specifies the point 4, 0.
Then we have 3i plus 3j.
So 1, 2, 3.
1, 2, 3.
That is that point right there.
The position vector specifies that point,
or specifies the point 3, 3.
And then finally, we have negative 3i.
Let me do this in another color.
Negative 3i.
So 1, 2, 3, plus 2j, 1, 2.
So right over there.
So that is that last position vector,
and it's specifying the point negative 3, 2.
And they want to know, what's the quadrilateral made
by these four positions?
So let's draw it.
Let's connect the dots.
So this is P, go to Q. It gets us that line right over there.
Then from Q to R, we're going to go like that.
And then from R to S, we're going to go like that.
Actually, it's going to go a little bit more--
it's going to go like that.
Sorry, that's from R to S. And then S to P
is going to be like that.
So just eyeballing it, it's clearly not a square.
It's clearly not a rectangle.
It's coming off at different angles.
It's clearly not a rhombus.
A rhombus would have all of the same sides.
So a square would be all 90 degrees,
would be all the same sides and all the angles are 90 degrees.
That would be a square.
It's clearly not a square.
A rectangle would be all the angles are 90 degrees,
but two sides are going to be the same,
and the other two sides are going to be the same.
These do not look like 90 degree angles.
And you could try if you want.
I mean, just eyeballing it, it's pretty
clear these aren't 90 degree angles.
But if you wanted to, you could find the slope of this
and find the slope of that line, and if they're not
the negative inverse of each other,
then they're not perpendicular, and they're not
coming at 90 degree angles.
And so the last one is are they a rhombus.
A rhombus is a parallelogram while the sides are the same.
So a rhombus would look like that.
Clearly not a rhombus.
These sides are clearly much longer
than those sides over there.
Now just from deductive reasoning,
it looks like the answer is A. it's a parallelogram.
But we can verify it.
In order for this to be a parallelogram,
these two sides have to be parallel,
and these two sides have to be parallel.
Or another way to think of it, they
would have to have the same slope.
So we could verify it.
If we were taking this on exam under time pressure,
we would just cross these out, and it looks pretty good.
We would just go with A. But let's just verify it.
So what's the slope down here?
So it's change in y over change in x.
So 0 minus negative 1, over 4 minus negative 2,
or this is equal to 1 over 4 minus negative 2,
is positive 6.
This is 1/6.
Or another way to think about it, we ran 6
and we only rised 1.
This is 1/6.
Now what happens over here?
Well, it's the same thing.
We're going from negative 3 to three.
So we're running 6.
So we say 3 minus negative 3 is equal to 6.
And how much did we rise?
We have 3 minus 2, which is equal to 1.
So this is also a slope of 1/6.
So these two lines are clearly parallel.
And now let's do it for these other two.
What's the slope over here?
And actually, you could just do rise over run.
So that's exactly what we were doing before,
but just to give you the taste for everything.
So we ran, to go from here to here,
we have to move to the right 1.
So we run 1.
And how much do we rise?
Well, we had to go down 3.
So it's negative 3.
So the slope here is negative 3.
And what about here?
To go from this point to this point, we went to the right 1.
So we ran 1.
And we had to go down-- we started at 2,
and we ended up at negative 1.
We had to go down 3.
So the slope here is also negative 3.
So these lines are also parallel.
So we're definitely dealing with a parallelogram, which
is neither a rhombus nor a rectangle.