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>> We're going to figure
out how to find the distance
between two point
on the plane,
and from that we will discover
the distance formula.
So here's an example.
Find the distance
between negative 3,5 and 6,-2.
Now, one way
of doing this is going ahead
and plotting these points
on the XY plane here.
So let's see.
Negative 3,5;
negative 3,5 is right here.
Okay. So that's the point
negative 3,5.
And then we've got 6,-2,
so that's over 6 and on 2
and that's the ordered pair 6,
2.
So what we're trying to find
out is the straight line
between those two point,
what's that exact distance.
So I'm looking
for this distance D.
All right.
Well, let's see.
How about if we try
to make a right triangle here.
All right.
And this is assuming you
already know the
Pythagorean theorem.
So you have two legs
of a triangle.
And if you take the length
of one side squared plus the
length of the other side
squared for the legs,
that's equal
to the hypotenuse squared.
And the distance,
that's really the hypotenuse
of this triangle.
So how long are each
of these sides?
Well, let's see.
I've got one 1, 2, 3, 4, 5, 6,
7 spaces here.
And along the X axis here
on the horizontal direction
I've got 3 plus an extra 6.
I've got 9 spaces there.
So if you use the Pythagorean
theorem, I could say
that the hypotenuse squared,
D squared, okay,
equals 7 squared plus
9 squared.
So D squared equals 49 plus
81, so D squared equals 130.
All right?
So that's the square
of the distance.
I want to know what D is,
so we take the square root
of both sides.
Now, when you normally take
the square root,
you take the plus
or minus sign in front
of the square root sign.
But, of course,
this is a distance,
so a distance can't
be negative.
So we would say
that distance is the square
root of 130.
And that's exactly how long
that side is.
It can't be simplified.
And you could get
out a calculator and round it.
So if you want
to know approximately how long
that is, you would get
out your calculator.
And to the nearest tenth is
about 11.4.
So, first of all,
the hypotenuse is always
longer than either
of the other side,
but it's always shorter
than the sum of the two sides,
right, because the long way
would be to walk 7 plus 9.
So this is a short distance.
So look at the picture,
and that seems reasonable
that the distance is about 11
and a half.
All right.
So the exact answer here is
square root of 130.
The approximate distance
rounded to the nearest tenth
is 11.4.
Now, if you want
to find the distance
between two points,
you don't always want to have
to get out a piece
of graph paper
and do it this way.
So it would be nice
if we had some sort of formula
to use, especially imagine
if these were really large
numbers like -300, 568,
you know, you don't want
to have to try to draw that.
One other point --
one other point
that you can write
down is this point here
at the right angle,
what ordered pair is that?
It's -3, all right,
and it's down -2.
So notice it's the X
coordinate for that point,
and it's the Y coordinate
for that point.
Now, what about the
distances here?
How would we get 9?
Well, look
at the X values here.
See this --
these two points are
up at the same height or down
at the same place if you want
to think about negative
two spaces.
But what is that distance?
Well, it's going
where X is -3 all the way
over to 6.
So we've got three places plus
6 more spaces.
That makes sense.
That's 9 spaces.
Could we use these X values
to get that answer?
It's really the larger number
of X minus the smaller number.
So notice that was just 6
minus the -3.
That's another way to get 9.
The same thing will happen
if I wanted
to get this Y value
of 7 right here.
How would we do
that without counting spaces?
The X value is the same.
But it's the Y values
that are different.
I've got the larger Y value
minus this smaller Y value,
5 minus -2.
Okay. So that's another way
to get the distance.
So if you know what the larger
number is,
the larger X value here,
along here you could take the
larger one minus the
smaller one.
If you're not sure,
like if you're working
with variables,
just keep in mind
that 6 minus -3,
if you take the absolute value
of it, it's the same thing
if you did negative 6 --
-3 minus 6.
In other words,
if you subtract the other way
around, as long
as you put the absolute value
signs around it,
it's going
to be the same thing.
It's important
to pay attention to.
So we're going to use some
of these ideas
to get a formula
if we are going
to use variables
for your points.
All right?
So, in other words,
if we have X1 Y1,
we're not sure where it is
in the plane, but we're going
to try to use the same ideas I
have here to go ahead
and use the Pythagorean
theorem and get a
distance formula.
So here's the problem.
Find the distance
between X1 Y1 and X2 Y2.
Well, the problem is I really
don't know where those are
in the plane.
But let's just say there's
some point X1 Y1
and then some other point
X2 Y2.
Now, keep in mind it really
might not be lower.
It might be higher
at this point.
You really don't know.
The X1 Y1 could be
to the right
of where the X2 Y2 is.
So I have to be very careful
to realize
that it might not look exactly
like this.
But our goal is
to find the distance
between those two points.
So I'm going
to make a right triangle
again, and I'm going to try
and figure
out what this ordered pair is
right here.
Okay. So if I was
in the XY plane,
how far over is this?
So what would the X value be
for this point here?
Well, it's
over the same distance
as this point up here,
the X1 Y1.
So the X coordinate has
to be X1 at that place.
And what's the Y value?
It's as high as that Y value
over here, Y2.
Now, the question is:
How long is this horizontal
segment here, okay?
Well the Y values are the
same, but there's a difference
with the X values.
You really don't know if X2 is
to the right of X1.
In this picture it is,
but I wouldn't know
in general.
But I know it's the difference
of the X and the Y values.
So I'm just going
to right it's the absolute
value of the difference.
Because that way it will be a
positive number,
and we know distances have
to be positive.
So if I don't put the absolute
value around it, I might end
up with a negative number.
All right.
How about this distance
over on the side here?
Well, that's the difference
of the Y values.
Again, from the picture,
it looks like Y1 is above Y2,
but it might not be
in general.
So, again,
it's going
to be the difference
of the Y values.
And I'm going
to put an absolute value
around it so it's clear
that it's a positive number.
Distance has to be positive.
Okay. Now, we're going
to use the
Pythagorean theorem.
We know that,
if you take the square
of one leg plus the square
of the other leg,
it equals the
distance squared.
So that's what I'm going
to write.
I'm going to write:
The distance squared is going
to be the square of one
of the legs.
Let's take this horizontal leg
here first.
So it's going
to be X1 minus X2.
Now, it's really an
absolute value.
But when I square it,
it will end
up being positive anyway.
So I could put an absolute
value around that X minus X2
and then square it.
But if X1 minus X2 was
negative, as long
as I square it,
I'm going to get the same
number here.
So you could check that out
for yourself.
If you do 3 minus 5
and then you square it,
you're going
to get the same thing
as if you do 5 minus 3
and then square
that number plus.
All right.
Now, that's this --
the length
of that leg squared plus the
length of the other
one squared.
And same reasoning.
I could put Y1 minus
Y2 squared.
So it's important
to notice it really doesn't
matter if I write X1 minus X2
or X2 minus X1.
I basically have
to subtract the X coordinates
and square it, add it to --
I have to subtract the Y
coordinates
and then square it.
Almost done.
To find the distance,
we just take the square root
of both sides.
So it's X1 minus X2 squared
plus Y1 minus Y2 squared.
Now, in algebra books,
you'll almost always see it
written like this:
D equals the square root
of X2 minus X1 squared plus Y2
minus Y1 squared.
But if really doesn't matter.
The trick is
in one parentheses you've got
to correct the X values
and square it,
and it doesn't matter
the order.
Because if you switch the
order, it just means it'll be
the opposite.
So instead
of being a positive 2,
it might be a negative 2.
Once you square it,
it will be the same number.
So as long
as you're squaring it,
it really doesn't matter
the order.
So these are exactly the
same formula.
And you should convince
yourself of that by putting
in some numbers if you want.
So let's go back
to that very first problem,
and now we're going
to use the distance formula.
So let's go back
to the original problem.
So this was our original
problem, find the distance
between negative 3,5 and 6,-2.
And when we did it before,
we found it was square root
of 130.
Now we're going
to use the formula.
Okay. So here's
where people say, well,
which one's X1
and which one's X2?
Well, it really
doesn't matter.
What matters is
that you're going
to subtract the X values.
I don't like working
with negative numbers.
It's easier for me to work
with positive numbers.
So here's how I set it up.
I put the square root,
put a parenthesis
with a minus sign
in the middle
and then square it,
add it to some other
parentheses
with a minus sign squared.
And now I'm going to fill
in the X values.
To make sure what is
in parentheses will be
positive, I just put the
bigger value of X first.
All right.
So the bigger value of X
in these two ordered pairs
is 6.
I'm going to put 6 before the
minus sign
and a negative 3 after it.
Notice if you did it the other
way, if you put a negative 3
first and then you did minus
6, you're going
to get negative 9;
and when you square it,
you're going to get 81,
which is what I'm going
to get, as well.
All right.
Now, let's take the other
ordered pairs, and we've got
to look at the two Y values.
And, again, just because I
like to have whatever's
in parentheses ending
up being a positive number,
I'm going to put the bigger
value of Y first, 5 and -2.
Now, you're going
to say, uh-uh.
That's really not the formula.
You are absolutely right.
The formula does say X1 minute
X2 and Y1 minus Y2,
but you can always switch
and write Y2 minus Y1.
It's not like the slope
formula where it makes a lot
of difference.
So if you put the -2 minus 5,
as long as you square it,
you're going
to get the same thing I do.
So this will be --
well, I have
to simplify inside
parentheses first.
That's order of operations.
So I've got 9 squared plus
7 squared.
Of course,
if you put in the values for X
and Y in a different order,
you would have maybe a
negative 9;
you'd have
to put maybe a parentheses
around that negative 9 before
you squared it.
So this is the square root
of 81 plus 49.
Still have
to add inside parentheses;
I'm sorry,
underneath the square root.
And you're still going
to get the square root of 130.
Now, if you want to take this
and fill in it exactly,
calling this first ordered
pair X1 X2,
the second order pair 6,-2;
so that's X1 Y1, X2 Y2,
plug it in exactly
as I've written it,
when all is said and done,
you should also get the square
root of 130.
We'll do some more problems
on another video.