-
In this unit, we're going to
have a look at two points.
-
And the line joining the two
-
points. We call this a straight
line segment and I'm going to
-
show you how we can calculate
the distance between the two
-
points, or In other words, the
length of this line segments and
-
how we can calculate the
coordinates of the midpoint of
-
the line segment.
-
Now, before we do that, I'd like
to revise some facts about the
-
coordinates of a point.
-
Suppose we choose a point, let's
call this .0.
-
And suppose we draw a horizontal
-
line. Through that point.
-
And a perpendicular line
align at 90 degrees also
-
through that point.
-
We call these two
perpendicular lines axes.
-
And this horizontal line. We'll
call it the X axis and label it
-
with the letter X.
-
And the vertical line.
-
Will call this the Y axis and
label it with the letter Y.
-
Now on these axes we draw a
scale and what we do is we
-
draw what's called a uniform
scale. That means it's evenly
-
spread along the axis.
-
And so on and on the Y axis we
do. Likewise, we put a uniform
-
scale on the Y axis.
-
And the point of doing that is
that we can then measure
-
distances in the X direction and
in the Y direction.
-
Now suppose we choose any
point at all in the plane.
-
Let's choose a point. Here,
let's call this point A.
-
What we can do is we can measure
the distance of a.
-
From the vertical axis, so we
-
can measure. That distance in
-
there. And because that's the
distance in the X direction,
-
we call it the X coordinate
of the point a.
-
We can also measure the distance
from the horizontal axis.
-
And because that's measured in
the Y direction, we call that
-
the Y coordinate of A.
-
And these two numbers, the X
coordinate and the Y coordinate
-
we write in brackets after the
letter A itself.
-
And we call XY the coordinates
of the point a.
-
We also refer to XY in the
brackets like. This is an
-
ordered pair of numbers, is
clearly a pair of numbers, and
-
it's an ordered pair because the
order is very important, we put
-
the X coordinate first. That's
the distance from the Y axis.
-
And the Y coordinate second,
that's the distance from the X
-
axis. So now we have a way
of specifying precisely
-
where in the plane our point
of interest lies.
-
If we move to the left hand side
of the Y axis, then what we have
-
are negative X coordinates
because as we move to the left.
-
We count down from zero, so
minus 1 - 2 - 3 and so on. So
-
anything to the left of the Y
axis will have a negative X
-
coordinate. And similarly, if
we want to come down below the
-
horizontal axis, we want to
come down here. We countdown
-
from 0 - 1 - 2 - 3 and so
on. So any point with a
-
negative Y coordinate will be
below the horizontal axis.
-
Let's plot some points
and see where they lie.
-
Not prepared, some
graph paper for us.
-
So here's my piece of graph
paper. You can see my X&Y Axis
-
and my uniform scales drawn on
the two axes, and let's plot
-
some points. Here's the first
Point, Point A and let's suppose
-
point a has coordinates 3. Two.
Let's see where that is.
-
Well, a has an X coordinate of
three, so we move in the X
-
Direction a distance 3.
-
And AY coordinate of two, so
we're moving up from the X axis
-
of distance to and that takes us
to this point here.
-
X coordinate of three Y
coordinate of two.
-
Let's have another point.
Let's have point B.
-
Switch those Round 2 three.
Let's see where that is now
-
with an X coordinate of two.
So we've got to move a
-
distance to from the Y axis.
-
And AY coordinate of three
means we move vertically. So
-
two across three up takes us
to this point there. And
-
that's my point, B.
-
Let's have some more. What
about this Point C, which has
-
an X coordinate of zero and a
Y coordinate of three? Let's
-
see where that is.
-
When X coordinate of 0 means we
move no distance at all in the X
-
direction, so we stay on the Y
-
axis. And we want to be a
distance three upwards from the
-
origin. So there's my Point C.
-
Point D that suppose that's
minus 3 zero. So I've introduced
-
a negative X cord in it. Now
remember with a negative X
-
coordinate, we're moving to the
left hand side of the Y axis. We
-
want to move a distance 3123, so
we're on this line here.
-
And AY coordinate is 0 means we
don't move up or down at all. So
-
at that point there. So there's
-
my point, D. Have a couple more
E minus 2 - 3 so we've got a
-
both a negative X and a negative
-
Y value. Minus 2 on the X
wings means we moved to the
-
left distance 2 - 3 on the Y
means we move down a distance
-
three, so we're down there to
get to point E.
-
Final one 2 - 2. Let's see
where that is.
-
X coordinate X coordinate of two
means. We moved to the writer.
-
Distance to. Y
coordinate of minus 2
-
means we moved down a
distance too, so
-
there's point F.
-
OK, so now given any ordered
pair of numbers representing a
-
point in the plane, you know
how to plot where it is. We
-
know how to locate it.
-
So now we know how to plot the
points. Let's see if we can plot
-
two points and find the distance
between the two points.
-
So some more graph paper here.
And let's suppose we pick any
-
two points and let me suppose
the first point, let's call it
-
a, has coordinates 13.
-
Let's put that on the graph.
-
X coordinate if one Y
coordinates of three means we
-
are one across and three up, so
at that point there that's a.
-
And let's choose my second
point to be four 5X
-
coordinate of four Y
coordinate of five puts me
-
there.
-
So two points.
-
Let's join them.
-
And that gives me my straight
line segment and what I'm going
-
to do is I'm going to show you
-
how we can calculate. The length
of this line segment, or In
-
other words, the distance
between points A&B.
-
Now, the way we do this is
as follows. We look at the
-
point A and will draw in a
horizontal line through a.
-
So because this line is
horizontal, it's parallel
-
to this X axis.
-
I'm going to draw in a
vertical line through be.
-
And again, because that's a
vertical line, it's parallel to
-
the Y axis, and so this angle in
here must be 90 degrees, and we
-
see what we've done is. We've
formed a right angle triangle.
-
Anywhere along this
horizontal line has a Y
-
coordinate of three, so in
particular the Y
-
coordinates at this point
here, where the right
-
angle is must be 3.
-
Call this Point C. It's Y
coordinate must be 3.
-
Now anywhere on this line.
-
Has an X coordinate of four the
same as the Point B, so the X
-
coordinate here must be 4. So
now we know the coordinates of
-
this point as well.
-
Let's put the coordinates of
being 45 and a in 1 three.
-
Now what I'm going to do is I'm
going to show you how we can
-
calculate this distance in here.
-
And this distance in here,
because then we will know the
-
lengths of two of the sides of
this right angle triangle.
-
And then we're going to use
Pythagoras theorem to get
-
this length, because this
length is the hypotenuse of
-
this right angle triangle.
-
Let's look at this
distance here.
-
Now the total distance from the
X axis up to point B.
-
Is 5. The Y coordinate of be.
-
So the distance all the way up
there is 5.
-
The distance from the X axis up
to Point C.
-
Is 3 the white coat?
-
You need to see so
that distance is 3.
-
So this distance that we're
really interested in here
-
must be the larger distance.
Subtract the smaller distance
-
5, subtract 3.
-
Which is 2.
-
It's important that you
recognize that we've calculated
-
this distance by finding the
difference of the Y coordinates,
-
5, subtract 3.
-
What about this distance
in here?
-
Well, the total distance
from the Y axis to this
-
dotted line is full.
-
Which is the X coordinate of be
or the X coordinate of C?
-
This shorter distance from
the Y axis to Point A is just
-
one. That's The X coordinate
of a, so that's one. So the
-
distance we're really
interested in is the larger
-
minus the smaller 4 - 1.
-
And 4 - 1 is 3.
-
And again 4 - 1 is the
difference of the two
-
X coordinates, 4 - 1,
giving you the three.
-
So now we know the length of the
base of this triangle is 3. We
-
know its height is too.
-
And we can use Pythagoras
theorem to find a B.
-
Pythagoras theorem will say
that AB squared.
-
Is the sum of the squares of the
other two sides. AB squared will
-
be 3 squared +2 squared.
-
3 squared is 9. Two squared is
four and nine and four is 30.
-
So AB squared is 13, so AB,
which is the distance we
-
want is going to be just the
square root of 13.
-
OK, so knowing the coordinates
of the two points.
-
We can find the lengths of the
sides of this triangle and then
-
use Pythagoras theorem to get
the hypotenuse, which is the
-
distance between the two points.
-
Now that's all well and good for
the specific case we looked at,
-
but what I want to do now is do
it more generally, instead of
-
having specific points, I want
to take two arbitrary points in
-
the plane and see how we can
find the distance between the
-
two of them. So again, let me
have a Y axis and X axis.
-
And and origin.
-
And let's pick our first point,
let's call it A and let's
-
suppose it's got an arbitrary X,
an arbitrary Y coordinate. So
-
let's call those X, one and Y
one. So there's my point A.
-
I'm going to take a second
point. Let's call that B.
-
And let's suppose its X
coordinate is X2, Y
-
coordinate is. Why two? So
these are arbitrary points.
-
Now could be anywhere.
-
We're going to try to find the
distance between the two
-
points, or In other words, the
length of this line segment.
-
As before, let's put
in a horizontal line.
-
Through a.
-
And a vertical line through be.
-
To form this right angle
-
triangle. Here, let's call
this Point C.
-
What are the coordinates
of points, see?
-
Well, anywhere along this
-
horizontal line. Will have a Y
coordinate, which is why one.
-
Because that's the same as the Y
coordinate of a. So the Y
-
coordinate of C must be why one?
-
What's its X coordinate? Well,
its X coordinate must be the
-
same as the X coordinate of bee
which is X2. So X2Y one are the
-
coordinates of Point C.
-
Let's calculate this
distance in here.
-
Now this distance.
-
Is the difference in the Y
coordinates the Y coordinates
-
of B subtract the Y coordinate
of C, so this is just Y 2
-
minus Y one. The difference in
the Y coordinates?
-
Similarly here this distance.
-
Is the X coordinate of C which
is X2 subtracted the X
-
coordinate of a which is X one.
So this distance is simply X2
-
minus X one.
-
And now you'll see, we know the
base and we know the height of
-
this right angle triangle and
we can use Pythagoras theorem
-
to get the hypotenuse, which is
the distance we require.
-
So Pythagoras theorem will say
that AB squared.
-
The square of the hypotenuse
is the sum of the squares of
-
the other two sides.
-
That's X2 minus X
one all squared.
-
Plus Y 2 minus
Y1 all squared.
-
And then if we want a B or we
have to do is square root both
-
sides, so maybe will be the
square root of X2 minus X one
-
all squared plus Y 2 minus Y one
-
or squared? And that is the
general formula that we can
-
always use to calculate the
distance between two arbitrary
-
points in the plane.
-
Let's look at a specific
example. Suppose we want to find
-
the distance between the points
AA, which has a coordinates
-
minus one and three and B which
-
has coordinates. Two and minus
four, for example, how can we
-
use this formula to find the
distance between these two
-
points without actually
drawing all the points?
-
Well, using the formula AB will
be the square root.
-
We want the distance, sorry. The
difference between the X
-
coordinates. So that's 2
minus minus one.
-
All squared.
-
Plus the difference
between the Y
-
coordinates, which is
minus 4 - 3.
-
All squared.
-
So that's the square root of 2.
Subtract minus one is 3 and 3
-
squared is 9.
-
Minus 4 - 3 is minus Seven and
minus 7 squared is plus 49.
-
So finally AB will be the square
root of 9 + 49.
-
Which is 58.
-
That's the distance between
those two points.
-
We're not going to look at how
you can calculate the
-
coordinates of the midpoint of a
straight line segment. Before we
-
do that, let me just.
-
Consider this problem. Suppose
we have an X axis.
-
And we have some points on the
X axis. So suppose we have one
-
2345, and so on.
-
Let's choose two points on this
X axis. Let's suppose we pick
-
this point. Where X is
2. Let's call that a.
-
And this point, where X is 4.
Let's call that be and ask
-
yourself what is the X value.
-
At the midpoint of the line
segment a be, so we're looking
-
at this segment Abe, and we want
to know what The X value is at
-
the midpoint. I think it's
fairly obvious from looking
-
at it that the X value at
the midpoint is 3.
-
Another way of thinking of that
is that 3.
-
Is the average of two and four.
-
Because if we average two
and four Adam together
-
divide by 2, two and four
is 6 over 2 which is 3.
-
So In other words, The X
coordinates at the midpoint
-
of a line segment is the
average of the X coordinates
-
at the two ends.
-
Just to convince you of
that, let's have a look at
-
another problem.
-
So we have one 23456
and Seven. And let's suppose
-
this time we pick, say,
the .3 for A.
-
And the point 74B.
-
And we're looking at this
straight line. Segments here
-
between A&B. Again, it's rather
obvious from looking at it.
-
The X value at the
midpoint is 5.
-
And if we average the X values
at the beginning and the end.
-
Of the line segment
three added to 7.
-
Divided by two as the average of
the X values is 10 over 2, which
-
is 5. So in a more general case,
in a few minutes time will see
-
that when we want to find the
coordinates of the midpoint of a
-
line, we just need to average
the values at either end.
-
OK, let's look at a more general
one. Suppose we have two points.
-
Suppose A is the .23, B is the
-
point. 45
-
Let's plot those. And then we're
going to try to find the
-
coordinates of the midpoint of
the line joining them.
-
OK. So as the .23,
so here's Point A.
-
And Point B is 45, so be will be
-
up here. And about that.
-
And there's my line segment.
-
Maybe?
-
We are interested in the
point which is the midpoint
-
of this line, so let's put
that on there.
-
And let's call it point P.
-
And because it's the
midpoint of the line, we
-
know that this length must
be the same as this length.
-
AP Is the same length as PB.
-
Let's put some other
lines in here.
-
A horizontal line through a.
-
Vertical line through be.
-
Horizontal line through P and a
vertical line through P.
-
And let's label some of these
points that we've created C.
-
And D. And he.
-
Now let's look at the line
segment from A to E.
-
OK.
-
Sorry Peter.
-
Not step.
-
OK.
-
Yeah.
-
Now let's look at the triangle.
-
ACP. And the
-
triangle BD. Sorry, the
triangle.
-
Yeah.
-
Now let's look at
the triangle ACP.
-
And the triangle PD.
-
Be.
-
Both of these triangles are
right angle triangles
-
because they both have a
right angle here.
-
Both of the triangles have
the same hypotenuse, which
-
has the same length 'cause
they both have the same
-
length there in there.
-
Also, this angle in here.
-
Is the same as this angle in
here because their corresponding
-
angles? So in fact, because
we've got two angles and a
-
side in the first triangle
corresponding to two angles
-
and the side in the second
triangle, these two triangles
-
or what we call congruent.
-
Congruent triangles are
identical in every respect, and
-
in particular. What this means
is that this length from B down
-
to D. Most corresponding
with the same as this length
-
from P down to see.
-
So In other words, because these
two lengths are the same BD and
-
PC, which is the same as DE.
-
Then this point D.
-
Must be the midpoint of the line
from A to B.
-
So the Y coordinate of D
must be the average of
-
the Y coordinates at B&B.
-
Now the average of the Y
coordinates at E&B is 3 + 5
-
/ 2, which is 8 over 2 which
is 4. So the Y coordinates
-
at D must be 4.
-
What about the X coordinate?
-
If we look at this line from A
-
to C. Because the triangles are
congruent, then this distance
-
from A to see must be the same
as the distance from P to D,
-
which is the same as the
distance from C to E.
-
So In other words, because these
two distances are the same, see
-
must be the midpoint of AE.
-
And because C is the midpoint of
-
AE. It's X coordinate must be
the average of the X coordinates
-
A&E. Now the average of those X
coordinates is 2 + 4, which is 6
-
/ 2 which is 3. So the X
coordinates here must be 3.
-
So the X coordinate is 3. We
know the Y coordinate is 3. At
-
this point we decided that the Y
coordinate was four and we know
-
the X coordinate is also 4.
-
So now we can read off what the
coordinates of point PRP will
-
have an X Coordinate A3.
-
And AY coordinate of four.
-
Now that's very complicated. It
seems very complicated, but in
-
practice it's very simple. Let's
just look at these numbers
-
again. The X coordinate at the
-
midpoint. Which is 3 is just the
average of the two X coordinates
-
2 + 46 / 2 is 3, the Y
coordinate at the mid .4 is just
-
the average of the two Y
coordinates. Five and three,
-
which is 8 / 2 is 4.
-
So in general, if we have
a situation where we've got a
-
point AX1Y one and point BX2Y2.
-
The coordinates of the midpoint
of the line AV are the average
-
of the X coordinates.
-
And the average of the Y
coordinates.
-
Let's just look at a
simple example just to
-
hammer that home.
-
Suppose we have a point A with
coordinates 2 - 4 and be with
-
coordinates minus four and
three. So two points and were
-
interested in the coordinates of
the midpoint of the line AB.
-
Select this point BP.
-
The X coordinate will be the
average of the two X
-
coordinates, which is 2. Added 2
- 4 over 2.
-
And the Y coordinate will be the
average of the two Y
-
coordinates, which is minus four
added to 3 / 2.
-
And if we tie do that up, will
get 2 subtract 4, which is minus
-
2 over 2, which is minus one.
-
And minus 4 + 3 is minus 1 - 1
over 2 is minus 1/2.
-
So there is the coordinates
of the midpoint of the line
-
joining amb.