[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:01.96,0:00:05.34,Default,,0000,0000,0000,,In this unit, we're going to\Nhave a look at two points.
Dialogue: 0,0:00:07.73,0:00:09.59,Default,,0000,0000,0000,,And the line joining the two
Dialogue: 0,0:00:09.59,0:00:14.53,Default,,0000,0000,0000,,points. We call this a straight\Nline segment and I'm going to
Dialogue: 0,0:00:14.53,0:00:18.60,Default,,0000,0000,0000,,show you how we can calculate\Nthe distance between the two
Dialogue: 0,0:00:18.60,0:00:23.04,Default,,0000,0000,0000,,points, or In other words, the\Nlength of this line segments and
Dialogue: 0,0:00:23.04,0:00:26.74,Default,,0000,0000,0000,,how we can calculate the\Ncoordinates of the midpoint of
Dialogue: 0,0:00:26.74,0:00:27.85,Default,,0000,0000,0000,,the line segment.
Dialogue: 0,0:00:28.54,0:00:32.26,Default,,0000,0000,0000,,Now, before we do that, I'd like\Nto revise some facts about the
Dialogue: 0,0:00:32.26,0:00:33.40,Default,,0000,0000,0000,,coordinates of a point.
Dialogue: 0,0:00:34.72,0:00:38.53,Default,,0000,0000,0000,,Suppose we choose a point, let's\Ncall this .0.
Dialogue: 0,0:00:39.79,0:00:42.23,Default,,0000,0000,0000,,And suppose we draw a horizontal
Dialogue: 0,0:00:42.23,0:00:44.26,Default,,0000,0000,0000,,line. Through that point.
Dialogue: 0,0:00:44.85,0:00:47.81,Default,,0000,0000,0000,,And a perpendicular line\Nalign at 90 degrees also
Dialogue: 0,0:00:47.81,0:00:48.80,Default,,0000,0000,0000,,through that point.
Dialogue: 0,0:00:51.06,0:00:54.43,Default,,0000,0000,0000,,We call these two\Nperpendicular lines axes.
Dialogue: 0,0:00:57.88,0:01:02.77,Default,,0000,0000,0000,,And this horizontal line. We'll\Ncall it the X axis and label it
Dialogue: 0,0:01:02.77,0:01:04.27,Default,,0000,0000,0000,,with the letter X.
Dialogue: 0,0:01:05.06,0:01:06.38,Default,,0000,0000,0000,,And the vertical line.
Dialogue: 0,0:01:07.70,0:01:11.76,Default,,0000,0000,0000,,Will call this the Y axis and\Nlabel it with the letter Y.
Dialogue: 0,0:01:13.46,0:01:17.77,Default,,0000,0000,0000,,Now on these axes we draw a\Nscale and what we do is we
Dialogue: 0,0:01:17.77,0:01:20.85,Default,,0000,0000,0000,,draw what's called a uniform\Nscale. That means it's evenly
Dialogue: 0,0:01:20.85,0:01:22.08,Default,,0000,0000,0000,,spread along the axis.
Dialogue: 0,0:01:25.58,0:01:30.17,Default,,0000,0000,0000,,And so on and on the Y axis we\Ndo. Likewise, we put a uniform
Dialogue: 0,0:01:30.17,0:01:31.70,Default,,0000,0000,0000,,scale on the Y axis.
Dialogue: 0,0:01:34.48,0:01:38.19,Default,,0000,0000,0000,,And the point of doing that is\Nthat we can then measure
Dialogue: 0,0:01:38.19,0:01:41.28,Default,,0000,0000,0000,,distances in the X direction and\Nin the Y direction.
Dialogue: 0,0:01:42.69,0:01:46.01,Default,,0000,0000,0000,,Now suppose we choose any\Npoint at all in the plane.
Dialogue: 0,0:01:46.01,0:01:49.03,Default,,0000,0000,0000,,Let's choose a point. Here,\Nlet's call this point A.
Dialogue: 0,0:01:50.47,0:01:54.36,Default,,0000,0000,0000,,What we can do is we can measure\Nthe distance of a.
Dialogue: 0,0:01:54.97,0:01:56.92,Default,,0000,0000,0000,,From the vertical axis, so we
Dialogue: 0,0:01:56.92,0:01:59.18,Default,,0000,0000,0000,,can measure. That distance in
Dialogue: 0,0:01:59.18,0:02:03.70,Default,,0000,0000,0000,,there. And because that's the\Ndistance in the X direction,
Dialogue: 0,0:02:03.70,0:02:07.46,Default,,0000,0000,0000,,we call it the X coordinate\Nof the point a.
Dialogue: 0,0:02:09.36,0:02:12.56,Default,,0000,0000,0000,,We can also measure the distance\Nfrom the horizontal axis.
Dialogue: 0,0:02:15.06,0:02:18.12,Default,,0000,0000,0000,,And because that's measured in\Nthe Y direction, we call that
Dialogue: 0,0:02:18.12,0:02:19.51,Default,,0000,0000,0000,,the Y coordinate of A.
Dialogue: 0,0:02:20.93,0:02:24.72,Default,,0000,0000,0000,,And these two numbers, the X\Ncoordinate and the Y coordinate
Dialogue: 0,0:02:24.72,0:02:27.83,Default,,0000,0000,0000,,we write in brackets after the\Nletter A itself.
Dialogue: 0,0:02:30.39,0:02:34.74,Default,,0000,0000,0000,,And we call XY the coordinates\Nof the point a.
Dialogue: 0,0:02:36.37,0:02:40.07,Default,,0000,0000,0000,,We also refer to XY in the\Nbrackets like. This is an
Dialogue: 0,0:02:40.07,0:02:43.45,Default,,0000,0000,0000,,ordered pair of numbers, is\Nclearly a pair of numbers, and
Dialogue: 0,0:02:43.45,0:02:47.15,Default,,0000,0000,0000,,it's an ordered pair because the\Norder is very important, we put
Dialogue: 0,0:02:47.15,0:02:50.54,Default,,0000,0000,0000,,the X coordinate first. That's\Nthe distance from the Y axis.
Dialogue: 0,0:02:51.20,0:02:55.02,Default,,0000,0000,0000,,And the Y coordinate second,\Nthat's the distance from the X
Dialogue: 0,0:02:55.02,0:02:59.24,Default,,0000,0000,0000,,axis. So now we have a way\Nof specifying precisely
Dialogue: 0,0:02:59.24,0:03:02.36,Default,,0000,0000,0000,,where in the plane our point\Nof interest lies.
Dialogue: 0,0:03:03.79,0:03:08.80,Default,,0000,0000,0000,,If we move to the left hand side\Nof the Y axis, then what we have
Dialogue: 0,0:03:08.80,0:03:12.24,Default,,0000,0000,0000,,are negative X coordinates\Nbecause as we move to the left.
Dialogue: 0,0:03:14.50,0:03:19.80,Default,,0000,0000,0000,,We count down from zero, so\Nminus 1 - 2 - 3 and so on. So
Dialogue: 0,0:03:19.80,0:03:24.10,Default,,0000,0000,0000,,anything to the left of the Y\Naxis will have a negative X
Dialogue: 0,0:03:24.10,0:03:29.08,Default,,0000,0000,0000,,coordinate. And similarly, if\Nwe want to come down below the
Dialogue: 0,0:03:29.08,0:03:32.78,Default,,0000,0000,0000,,horizontal axis, we want to\Ncome down here. We countdown
Dialogue: 0,0:03:32.78,0:03:38.70,Default,,0000,0000,0000,,from 0 - 1 - 2 - 3 and so\Non. So any point with a
Dialogue: 0,0:03:38.70,0:03:42.03,Default,,0000,0000,0000,,negative Y coordinate will be\Nbelow the horizontal axis.
Dialogue: 0,0:03:43.09,0:03:45.31,Default,,0000,0000,0000,,Let's plot some points\Nand see where they lie.
Dialogue: 0,0:03:49.46,0:03:51.50,Default,,0000,0000,0000,,Not prepared, some\Ngraph paper for us.
Dialogue: 0,0:03:55.77,0:04:00.15,Default,,0000,0000,0000,,So here's my piece of graph\Npaper. You can see my X&Y Axis
Dialogue: 0,0:04:00.15,0:04:04.20,Default,,0000,0000,0000,,and my uniform scales drawn on\Nthe two axes, and let's plot
Dialogue: 0,0:04:04.20,0:04:07.90,Default,,0000,0000,0000,,some points. Here's the first\NPoint, Point A and let's suppose
Dialogue: 0,0:04:07.90,0:04:11.61,Default,,0000,0000,0000,,point a has coordinates 3. Two.\NLet's see where that is.
Dialogue: 0,0:04:12.89,0:04:17.92,Default,,0000,0000,0000,,Well, a has an X coordinate of\Nthree, so we move in the X
Dialogue: 0,0:04:17.92,0:04:19.35,Default,,0000,0000,0000,,Direction a distance 3.
Dialogue: 0,0:04:21.51,0:04:25.27,Default,,0000,0000,0000,,And AY coordinate of two, so\Nwe're moving up from the X axis
Dialogue: 0,0:04:25.27,0:04:28.45,Default,,0000,0000,0000,,of distance to and that takes us\Nto this point here.
Dialogue: 0,0:04:31.02,0:04:34.18,Default,,0000,0000,0000,,X coordinate of three Y\Ncoordinate of two.
Dialogue: 0,0:04:35.69,0:04:38.19,Default,,0000,0000,0000,,Let's have another point.\NLet's have point B.
Dialogue: 0,0:04:39.33,0:04:42.67,Default,,0000,0000,0000,,Switch those Round 2 three.\NLet's see where that is now
Dialogue: 0,0:04:42.67,0:04:46.32,Default,,0000,0000,0000,,with an X coordinate of two.\NSo we've got to move a
Dialogue: 0,0:04:46.32,0:04:48.15,Default,,0000,0000,0000,,distance to from the Y axis.
Dialogue: 0,0:04:49.60,0:04:53.04,Default,,0000,0000,0000,,And AY coordinate of three\Nmeans we move vertically. So
Dialogue: 0,0:04:53.04,0:04:56.82,Default,,0000,0000,0000,,two across three up takes us\Nto this point there. And
Dialogue: 0,0:04:56.82,0:04:58.20,Default,,0000,0000,0000,,that's my point, B.
Dialogue: 0,0:05:00.00,0:05:03.12,Default,,0000,0000,0000,,Let's have some more. What\Nabout this Point C, which has
Dialogue: 0,0:05:03.12,0:05:06.53,Default,,0000,0000,0000,,an X coordinate of zero and a\NY coordinate of three? Let's
Dialogue: 0,0:05:06.53,0:05:07.67,Default,,0000,0000,0000,,see where that is.
Dialogue: 0,0:05:08.77,0:05:12.91,Default,,0000,0000,0000,,When X coordinate of 0 means we\Nmove no distance at all in the X
Dialogue: 0,0:05:12.91,0:05:14.84,Default,,0000,0000,0000,,direction, so we stay on the Y
Dialogue: 0,0:05:14.84,0:05:19.92,Default,,0000,0000,0000,,axis. And we want to be a\Ndistance three upwards from the
Dialogue: 0,0:05:19.92,0:05:22.11,Default,,0000,0000,0000,,origin. So there's my Point C.
Dialogue: 0,0:05:25.84,0:05:29.43,Default,,0000,0000,0000,,Point D that suppose that's\Nminus 3 zero. So I've introduced
Dialogue: 0,0:05:29.43,0:05:33.34,Default,,0000,0000,0000,,a negative X cord in it. Now\Nremember with a negative X
Dialogue: 0,0:05:33.34,0:05:37.58,Default,,0000,0000,0000,,coordinate, we're moving to the\Nleft hand side of the Y axis. We
Dialogue: 0,0:05:37.58,0:05:41.49,Default,,0000,0000,0000,,want to move a distance 3123, so\Nwe're on this line here.
Dialogue: 0,0:05:43.44,0:05:47.84,Default,,0000,0000,0000,,And AY coordinate is 0 means we\Ndon't move up or down at all. So
Dialogue: 0,0:05:47.84,0:05:49.59,Default,,0000,0000,0000,,at that point there. So there's
Dialogue: 0,0:05:49.59,0:05:56.20,Default,,0000,0000,0000,,my point, D. Have a couple more\NE minus 2 - 3 so we've got a
Dialogue: 0,0:05:56.20,0:05:58.80,Default,,0000,0000,0000,,both a negative X and a negative
Dialogue: 0,0:05:58.80,0:06:04.00,Default,,0000,0000,0000,,Y value. Minus 2 on the X\Nwings means we moved to the
Dialogue: 0,0:06:04.00,0:06:08.48,Default,,0000,0000,0000,,left distance 2 - 3 on the Y\Nmeans we move down a distance
Dialogue: 0,0:06:08.48,0:06:11.68,Default,,0000,0000,0000,,three, so we're down there to\Nget to point E.
Dialogue: 0,0:06:13.75,0:06:18.20,Default,,0000,0000,0000,,Final one 2 - 2. Let's see\Nwhere that is.
Dialogue: 0,0:06:19.41,0:06:23.03,Default,,0000,0000,0000,,X coordinate X coordinate of two\Nmeans. We moved to the writer.
Dialogue: 0,0:06:23.03,0:06:26.44,Default,,0000,0000,0000,,Distance to. Y\Ncoordinate of minus 2
Dialogue: 0,0:06:26.44,0:06:29.42,Default,,0000,0000,0000,,means we moved down a\Ndistance too, so
Dialogue: 0,0:06:29.42,0:06:30.54,Default,,0000,0000,0000,,there's point F.
Dialogue: 0,0:06:31.88,0:06:34.99,Default,,0000,0000,0000,,OK, so now given any ordered\Npair of numbers representing a
Dialogue: 0,0:06:34.99,0:06:38.67,Default,,0000,0000,0000,,point in the plane, you know\Nhow to plot where it is. We
Dialogue: 0,0:06:38.67,0:06:40.09,Default,,0000,0000,0000,,know how to locate it.
Dialogue: 0,0:06:47.73,0:06:51.93,Default,,0000,0000,0000,,So now we know how to plot the\Npoints. Let's see if we can plot
Dialogue: 0,0:06:51.93,0:06:54.73,Default,,0000,0000,0000,,two points and find the distance\Nbetween the two points.
Dialogue: 0,0:07:01.81,0:07:06.71,Default,,0000,0000,0000,,So some more graph paper here.\NAnd let's suppose we pick any
Dialogue: 0,0:07:06.71,0:07:11.60,Default,,0000,0000,0000,,two points and let me suppose\Nthe first point, let's call it
Dialogue: 0,0:07:11.60,0:07:13.23,Default,,0000,0000,0000,,a, has coordinates 13.
Dialogue: 0,0:07:15.51,0:07:16.80,Default,,0000,0000,0000,,Let's put that on the graph.
Dialogue: 0,0:07:17.98,0:07:21.23,Default,,0000,0000,0000,,X coordinate if one Y\Ncoordinates of three means we
Dialogue: 0,0:07:21.23,0:07:25.46,Default,,0000,0000,0000,,are one across and three up, so\Nat that point there that's a.
Dialogue: 0,0:07:26.84,0:07:31.14,Default,,0000,0000,0000,,And let's choose my second\Npoint to be four 5X
Dialogue: 0,0:07:31.14,0:07:35.01,Default,,0000,0000,0000,,coordinate of four Y\Ncoordinate of five puts me
Dialogue: 0,0:07:35.01,0:07:35.44,Default,,0000,0000,0000,,there.
Dialogue: 0,0:07:37.64,0:07:38.68,Default,,0000,0000,0000,,So two points.
Dialogue: 0,0:07:39.80,0:07:40.70,Default,,0000,0000,0000,,Let's join them.
Dialogue: 0,0:07:44.05,0:07:46.71,Default,,0000,0000,0000,,And that gives me my straight\Nline segment and what I'm going
Dialogue: 0,0:07:46.71,0:07:48.49,Default,,0000,0000,0000,,to do is I'm going to show you
Dialogue: 0,0:07:48.49,0:07:52.77,Default,,0000,0000,0000,,how we can calculate. The length\Nof this line segment, or In
Dialogue: 0,0:07:52.77,0:07:55.15,Default,,0000,0000,0000,,other words, the distance\Nbetween points A&B.
Dialogue: 0,0:07:56.62,0:08:00.78,Default,,0000,0000,0000,,Now, the way we do this is\Nas follows. We look at the
Dialogue: 0,0:08:00.78,0:08:04.30,Default,,0000,0000,0000,,point A and will draw in a\Nhorizontal line through a.
Dialogue: 0,0:08:08.25,0:08:10.79,Default,,0000,0000,0000,,So because this line is\Nhorizontal, it's parallel
Dialogue: 0,0:08:10.79,0:08:12.07,Default,,0000,0000,0000,,to this X axis.
Dialogue: 0,0:08:13.28,0:08:16.25,Default,,0000,0000,0000,,I'm going to draw in a\Nvertical line through be.
Dialogue: 0,0:08:21.02,0:08:24.25,Default,,0000,0000,0000,,And again, because that's a\Nvertical line, it's parallel to
Dialogue: 0,0:08:24.25,0:08:29.10,Default,,0000,0000,0000,,the Y axis, and so this angle in\Nhere must be 90 degrees, and we
Dialogue: 0,0:08:29.10,0:08:32.65,Default,,0000,0000,0000,,see what we've done is. We've\Nformed a right angle triangle.
Dialogue: 0,0:08:34.87,0:08:37.92,Default,,0000,0000,0000,,Anywhere along this\Nhorizontal line has a Y
Dialogue: 0,0:08:37.92,0:08:40.97,Default,,0000,0000,0000,,coordinate of three, so in\Nparticular the Y
Dialogue: 0,0:08:40.97,0:08:44.01,Default,,0000,0000,0000,,coordinates at this point\Nhere, where the right
Dialogue: 0,0:08:44.01,0:08:45.92,Default,,0000,0000,0000,,angle is must be 3.
Dialogue: 0,0:08:47.34,0:08:50.87,Default,,0000,0000,0000,,Call this Point C. It's Y\Ncoordinate must be 3.
Dialogue: 0,0:08:52.55,0:08:54.07,Default,,0000,0000,0000,,Now anywhere on this line.
Dialogue: 0,0:08:55.38,0:08:59.73,Default,,0000,0000,0000,,Has an X coordinate of four the\Nsame as the Point B, so the X
Dialogue: 0,0:08:59.73,0:09:03.21,Default,,0000,0000,0000,,coordinate here must be 4. So\Nnow we know the coordinates of
Dialogue: 0,0:09:03.21,0:09:04.37,Default,,0000,0000,0000,,this point as well.
Dialogue: 0,0:09:05.67,0:09:11.78,Default,,0000,0000,0000,,Let's put the coordinates of\Nbeing 45 and a in 1 three.
Dialogue: 0,0:09:13.82,0:09:17.14,Default,,0000,0000,0000,,Now what I'm going to do is I'm\Ngoing to show you how we can
Dialogue: 0,0:09:17.14,0:09:18.24,Default,,0000,0000,0000,,calculate this distance in here.
Dialogue: 0,0:09:19.99,0:09:23.97,Default,,0000,0000,0000,,And this distance in here,\Nbecause then we will know the
Dialogue: 0,0:09:23.97,0:09:27.95,Default,,0000,0000,0000,,lengths of two of the sides of\Nthis right angle triangle.
Dialogue: 0,0:09:29.49,0:09:32.31,Default,,0000,0000,0000,,And then we're going to use\NPythagoras theorem to get
Dialogue: 0,0:09:32.31,0:09:34.85,Default,,0000,0000,0000,,this length, because this\Nlength is the hypotenuse of
Dialogue: 0,0:09:34.85,0:09:35.98,Default,,0000,0000,0000,,this right angle triangle.
Dialogue: 0,0:09:38.07,0:09:39.44,Default,,0000,0000,0000,,Let's look at this\Ndistance here.
Dialogue: 0,0:09:41.19,0:09:44.68,Default,,0000,0000,0000,,Now the total distance from the\NX axis up to point B.
Dialogue: 0,0:09:45.71,0:09:48.55,Default,,0000,0000,0000,,Is 5. The Y coordinate of be.
Dialogue: 0,0:09:49.48,0:09:52.86,Default,,0000,0000,0000,,So the distance all the way up\Nthere is 5.
Dialogue: 0,0:09:54.16,0:09:57.57,Default,,0000,0000,0000,,The distance from the X axis up\Nto Point C.
Dialogue: 0,0:09:58.30,0:10:00.04,Default,,0000,0000,0000,,Is 3 the white coat?
Dialogue: 0,0:10:00.11,0:10:03.20,Default,,0000,0000,0000,,You need to see so\Nthat distance is 3.
Dialogue: 0,0:10:04.43,0:10:07.35,Default,,0000,0000,0000,,So this distance that we're\Nreally interested in here
Dialogue: 0,0:10:07.35,0:10:10.26,Default,,0000,0000,0000,,must be the larger distance.\NSubtract the smaller distance
Dialogue: 0,0:10:10.26,0:10:11.23,Default,,0000,0000,0000,,5, subtract 3.
Dialogue: 0,0:10:13.23,0:10:13.96,Default,,0000,0000,0000,,Which is 2.
Dialogue: 0,0:10:15.38,0:10:18.08,Default,,0000,0000,0000,,It's important that you\Nrecognize that we've calculated
Dialogue: 0,0:10:18.08,0:10:21.46,Default,,0000,0000,0000,,this distance by finding the\Ndifference of the Y coordinates,
Dialogue: 0,0:10:21.46,0:10:22.48,Default,,0000,0000,0000,,5, subtract 3.
Dialogue: 0,0:10:24.22,0:10:25.93,Default,,0000,0000,0000,,What about this distance\Nin here?
Dialogue: 0,0:10:27.60,0:10:30.90,Default,,0000,0000,0000,,Well, the total distance\Nfrom the Y axis to this
Dialogue: 0,0:10:30.90,0:10:32.22,Default,,0000,0000,0000,,dotted line is full.
Dialogue: 0,0:10:33.39,0:10:36.68,Default,,0000,0000,0000,,Which is the X coordinate of be\Nor the X coordinate of C?
Dialogue: 0,0:10:39.33,0:10:43.52,Default,,0000,0000,0000,,This shorter distance from\Nthe Y axis to Point A is just
Dialogue: 0,0:10:43.52,0:10:47.71,Default,,0000,0000,0000,,one. That's The X coordinate\Nof a, so that's one. So the
Dialogue: 0,0:10:47.71,0:10:50.50,Default,,0000,0000,0000,,distance we're really\Ninterested in is the larger
Dialogue: 0,0:10:50.50,0:10:52.59,Default,,0000,0000,0000,,minus the smaller 4 - 1.
Dialogue: 0,0:10:54.08,0:10:55.57,Default,,0000,0000,0000,,And 4 - 1 is 3.
Dialogue: 0,0:10:56.52,0:11:00.10,Default,,0000,0000,0000,,And again 4 - 1 is the\Ndifference of the two
Dialogue: 0,0:11:00.10,0:11:03.02,Default,,0000,0000,0000,,X coordinates, 4 - 1,\Ngiving you the three.
Dialogue: 0,0:11:05.49,0:11:09.28,Default,,0000,0000,0000,,So now we know the length of the\Nbase of this triangle is 3. We
Dialogue: 0,0:11:09.28,0:11:10.55,Default,,0000,0000,0000,,know its height is too.
Dialogue: 0,0:11:11.51,0:11:15.00,Default,,0000,0000,0000,,And we can use Pythagoras\Ntheorem to find a B.
Dialogue: 0,0:11:16.39,0:11:19.75,Default,,0000,0000,0000,,Pythagoras theorem will say\Nthat AB squared.
Dialogue: 0,0:11:21.39,0:11:26.78,Default,,0000,0000,0000,,Is the sum of the squares of the\Nother two sides. AB squared will
Dialogue: 0,0:11:26.78,0:11:28.70,Default,,0000,0000,0000,,be 3 squared +2 squared.
Dialogue: 0,0:11:30.91,0:11:35.95,Default,,0000,0000,0000,,3 squared is 9. Two squared is\Nfour and nine and four is 30.
Dialogue: 0,0:11:37.15,0:11:41.85,Default,,0000,0000,0000,,So AB squared is 13, so AB,\Nwhich is the distance we
Dialogue: 0,0:11:41.85,0:11:46.17,Default,,0000,0000,0000,,want is going to be just the\Nsquare root of 13.
Dialogue: 0,0:11:47.66,0:11:50.28,Default,,0000,0000,0000,,OK, so knowing the coordinates\Nof the two points.
Dialogue: 0,0:11:51.11,0:11:55.04,Default,,0000,0000,0000,,We can find the lengths of the\Nsides of this triangle and then
Dialogue: 0,0:11:55.04,0:11:58.06,Default,,0000,0000,0000,,use Pythagoras theorem to get\Nthe hypotenuse, which is the
Dialogue: 0,0:11:58.06,0:11:59.57,Default,,0000,0000,0000,,distance between the two points.
Dialogue: 0,0:12:08.34,0:12:11.73,Default,,0000,0000,0000,,Now that's all well and good for\Nthe specific case we looked at,
Dialogue: 0,0:12:11.73,0:12:15.39,Default,,0000,0000,0000,,but what I want to do now is do\Nit more generally, instead of
Dialogue: 0,0:12:15.39,0:12:18.26,Default,,0000,0000,0000,,having specific points, I want\Nto take two arbitrary points in
Dialogue: 0,0:12:18.26,0:12:21.39,Default,,0000,0000,0000,,the plane and see how we can\Nfind the distance between the
Dialogue: 0,0:12:21.39,0:12:25.04,Default,,0000,0000,0000,,two of them. So again, let me\Nhave a Y axis and X axis.
Dialogue: 0,0:12:28.90,0:12:29.70,Default,,0000,0000,0000,,And and origin.
Dialogue: 0,0:12:30.91,0:12:35.16,Default,,0000,0000,0000,,And let's pick our first point,\Nlet's call it A and let's
Dialogue: 0,0:12:35.16,0:12:39.05,Default,,0000,0000,0000,,suppose it's got an arbitrary X,\Nan arbitrary Y coordinate. So
Dialogue: 0,0:12:39.05,0:12:43.65,Default,,0000,0000,0000,,let's call those X, one and Y\None. So there's my point A.
Dialogue: 0,0:12:45.26,0:12:47.93,Default,,0000,0000,0000,,I'm going to take a second\Npoint. Let's call that B.
Dialogue: 0,0:12:50.21,0:12:53.32,Default,,0000,0000,0000,,And let's suppose its X\Ncoordinate is X2, Y
Dialogue: 0,0:12:53.32,0:12:56.44,Default,,0000,0000,0000,,coordinate is. Why two? So\Nthese are arbitrary points.
Dialogue: 0,0:12:56.44,0:12:57.82,Default,,0000,0000,0000,,Now could be anywhere.
Dialogue: 0,0:13:00.25,0:13:03.67,Default,,0000,0000,0000,,We're going to try to find the\Ndistance between the two
Dialogue: 0,0:13:03.67,0:13:07.09,Default,,0000,0000,0000,,points, or In other words, the\Nlength of this line segment.
Dialogue: 0,0:13:09.02,0:13:11.74,Default,,0000,0000,0000,,As before, let's put\Nin a horizontal line.
Dialogue: 0,0:13:13.18,0:13:13.93,Default,,0000,0000,0000,,Through a.
Dialogue: 0,0:13:15.70,0:13:17.76,Default,,0000,0000,0000,,And a vertical line through be.
Dialogue: 0,0:13:23.19,0:13:24.87,Default,,0000,0000,0000,,To form this right angle
Dialogue: 0,0:13:24.87,0:13:27.83,Default,,0000,0000,0000,,triangle. Here, let's call\Nthis Point C.
Dialogue: 0,0:13:29.16,0:13:30.88,Default,,0000,0000,0000,,What are the coordinates\Nof points, see?
Dialogue: 0,0:13:32.44,0:13:34.21,Default,,0000,0000,0000,,Well, anywhere along this
Dialogue: 0,0:13:34.21,0:13:39.10,Default,,0000,0000,0000,,horizontal line. Will have a Y\Ncoordinate, which is why one.
Dialogue: 0,0:13:40.18,0:13:44.37,Default,,0000,0000,0000,,Because that's the same as the Y\Ncoordinate of a. So the Y
Dialogue: 0,0:13:44.37,0:13:46.62,Default,,0000,0000,0000,,coordinate of C must be why one?
Dialogue: 0,0:13:48.11,0:13:51.95,Default,,0000,0000,0000,,What's its X coordinate? Well,\Nits X coordinate must be the
Dialogue: 0,0:13:51.95,0:13:57.18,Default,,0000,0000,0000,,same as the X coordinate of bee\Nwhich is X2. So X2Y one are the
Dialogue: 0,0:13:57.18,0:13:58.58,Default,,0000,0000,0000,,coordinates of Point C.
Dialogue: 0,0:14:00.54,0:14:02.38,Default,,0000,0000,0000,,Let's calculate this\Ndistance in here.
Dialogue: 0,0:14:03.93,0:14:05.02,Default,,0000,0000,0000,,Now this distance.
Dialogue: 0,0:14:05.59,0:14:09.11,Default,,0000,0000,0000,,Is the difference in the Y\Ncoordinates the Y coordinates
Dialogue: 0,0:14:09.11,0:14:14.04,Default,,0000,0000,0000,,of B subtract the Y coordinate\Nof C, so this is just Y 2
Dialogue: 0,0:14:14.04,0:14:17.21,Default,,0000,0000,0000,,minus Y one. The difference in\Nthe Y coordinates?
Dialogue: 0,0:14:19.27,0:14:21.34,Default,,0000,0000,0000,,Similarly here this distance.
Dialogue: 0,0:14:22.71,0:14:27.10,Default,,0000,0000,0000,,Is the X coordinate of C which\Nis X2 subtracted the X
Dialogue: 0,0:14:27.10,0:14:31.86,Default,,0000,0000,0000,,coordinate of a which is X one.\NSo this distance is simply X2
Dialogue: 0,0:14:31.86,0:14:32.96,Default,,0000,0000,0000,,minus X one.
Dialogue: 0,0:14:34.32,0:14:38.95,Default,,0000,0000,0000,,And now you'll see, we know the\Nbase and we know the height of
Dialogue: 0,0:14:38.95,0:14:42.26,Default,,0000,0000,0000,,this right angle triangle and\Nwe can use Pythagoras theorem
Dialogue: 0,0:14:42.26,0:14:45.57,Default,,0000,0000,0000,,to get the hypotenuse, which is\Nthe distance we require.
Dialogue: 0,0:14:47.20,0:14:50.57,Default,,0000,0000,0000,,So Pythagoras theorem will say\Nthat AB squared.
Dialogue: 0,0:14:52.28,0:14:55.36,Default,,0000,0000,0000,,The square of the hypotenuse\Nis the sum of the squares of
Dialogue: 0,0:14:55.36,0:14:56.39,Default,,0000,0000,0000,,the other two sides.
Dialogue: 0,0:14:57.48,0:15:00.59,Default,,0000,0000,0000,,That's X2 minus X\None all squared.
Dialogue: 0,0:15:01.89,0:15:05.95,Default,,0000,0000,0000,,Plus Y 2 minus\NY1 all squared.
Dialogue: 0,0:15:08.11,0:15:14.67,Default,,0000,0000,0000,,And then if we want a B or we\Nhave to do is square root both
Dialogue: 0,0:15:14.67,0:15:20.00,Default,,0000,0000,0000,,sides, so maybe will be the\Nsquare root of X2 minus X one
Dialogue: 0,0:15:20.00,0:15:23.28,Default,,0000,0000,0000,,all squared plus Y 2 minus Y one
Dialogue: 0,0:15:23.28,0:15:28.64,Default,,0000,0000,0000,,or squared? And that is the\Ngeneral formula that we can
Dialogue: 0,0:15:28.64,0:15:32.31,Default,,0000,0000,0000,,always use to calculate the\Ndistance between two arbitrary
Dialogue: 0,0:15:32.31,0:15:33.95,Default,,0000,0000,0000,,points in the plane.
Dialogue: 0,0:15:35.02,0:15:38.77,Default,,0000,0000,0000,,Let's look at a specific\Nexample. Suppose we want to find
Dialogue: 0,0:15:38.77,0:15:42.18,Default,,0000,0000,0000,,the distance between the points\NAA, which has a coordinates
Dialogue: 0,0:15:42.18,0:15:44.57,Default,,0000,0000,0000,,minus one and three and B which
Dialogue: 0,0:15:44.57,0:15:48.48,Default,,0000,0000,0000,,has coordinates. Two and minus\Nfour, for example, how can we
Dialogue: 0,0:15:48.48,0:15:51.22,Default,,0000,0000,0000,,use this formula to find the\Ndistance between these two
Dialogue: 0,0:15:51.22,0:15:53.15,Default,,0000,0000,0000,,points without actually\Ndrawing all the points?
Dialogue: 0,0:15:54.57,0:15:58.72,Default,,0000,0000,0000,,Well, using the formula AB will\Nbe the square root.
Dialogue: 0,0:15:59.28,0:16:03.02,Default,,0000,0000,0000,,We want the distance, sorry. The\Ndifference between the X
Dialogue: 0,0:16:03.02,0:16:06.49,Default,,0000,0000,0000,,coordinates. So that's 2\Nminus minus one.
Dialogue: 0,0:16:08.92,0:16:09.72,Default,,0000,0000,0000,,All squared.
Dialogue: 0,0:16:11.62,0:16:14.97,Default,,0000,0000,0000,,Plus the difference\Nbetween the Y
Dialogue: 0,0:16:14.97,0:16:17.69,Default,,0000,0000,0000,,coordinates, which is\Nminus 4 - 3.
Dialogue: 0,0:16:19.95,0:16:20.70,Default,,0000,0000,0000,,All squared.
Dialogue: 0,0:16:22.51,0:16:27.47,Default,,0000,0000,0000,,So that's the square root of 2.\NSubtract minus one is 3 and 3
Dialogue: 0,0:16:27.47,0:16:28.53,Default,,0000,0000,0000,,squared is 9.
Dialogue: 0,0:16:29.67,0:16:35.41,Default,,0000,0000,0000,,Minus 4 - 3 is minus Seven and\Nminus 7 squared is plus 49.
Dialogue: 0,0:16:36.84,0:16:40.86,Default,,0000,0000,0000,,So finally AB will be the square\Nroot of 9 + 49.
Dialogue: 0,0:16:42.31,0:16:43.16,Default,,0000,0000,0000,,Which is 58.
Dialogue: 0,0:16:44.96,0:16:47.37,Default,,0000,0000,0000,,That's the distance between\Nthose two points.
Dialogue: 0,0:16:53.74,0:16:57.28,Default,,0000,0000,0000,,We're not going to look at how\Nyou can calculate the
Dialogue: 0,0:16:57.28,0:17:00.82,Default,,0000,0000,0000,,coordinates of the midpoint of a\Nstraight line segment. Before we
Dialogue: 0,0:17:00.82,0:17:02.43,Default,,0000,0000,0000,,do that, let me just.
Dialogue: 0,0:17:03.21,0:17:06.49,Default,,0000,0000,0000,,Consider this problem. Suppose\Nwe have an X axis.
Dialogue: 0,0:17:07.28,0:17:15.02,Default,,0000,0000,0000,,And we have some points on the\NX axis. So suppose we have one
Dialogue: 0,0:17:15.02,0:17:17.23,Default,,0000,0000,0000,,2345, and so on.
Dialogue: 0,0:17:18.33,0:17:22.16,Default,,0000,0000,0000,,Let's choose two points on this\NX axis. Let's suppose we pick
Dialogue: 0,0:17:22.16,0:17:25.99,Default,,0000,0000,0000,,this point. Where X is\N2. Let's call that a.
Dialogue: 0,0:17:27.17,0:17:32.16,Default,,0000,0000,0000,,And this point, where X is 4.\NLet's call that be and ask
Dialogue: 0,0:17:32.16,0:17:34.47,Default,,0000,0000,0000,,yourself what is the X value.
Dialogue: 0,0:17:35.12,0:17:38.34,Default,,0000,0000,0000,,At the midpoint of the line\Nsegment a be, so we're looking
Dialogue: 0,0:17:38.34,0:17:42.36,Default,,0000,0000,0000,,at this segment Abe, and we want\Nto know what The X value is at
Dialogue: 0,0:17:42.36,0:17:45.56,Default,,0000,0000,0000,,the midpoint. I think it's\Nfairly obvious from looking
Dialogue: 0,0:17:45.56,0:17:48.46,Default,,0000,0000,0000,,at it that the X value at\Nthe midpoint is 3.
Dialogue: 0,0:17:49.51,0:17:52.54,Default,,0000,0000,0000,,Another way of thinking of that\Nis that 3.
Dialogue: 0,0:17:53.17,0:17:55.44,Default,,0000,0000,0000,,Is the average of two and four.
Dialogue: 0,0:17:56.37,0:18:00.21,Default,,0000,0000,0000,,Because if we average two\Nand four Adam together
Dialogue: 0,0:18:00.21,0:18:05.76,Default,,0000,0000,0000,,divide by 2, two and four\Nis 6 over 2 which is 3.
Dialogue: 0,0:18:07.22,0:18:10.65,Default,,0000,0000,0000,,So In other words, The X\Ncoordinates at the midpoint
Dialogue: 0,0:18:10.65,0:18:14.42,Default,,0000,0000,0000,,of a line segment is the\Naverage of the X coordinates
Dialogue: 0,0:18:14.42,0:18:15.80,Default,,0000,0000,0000,,at the two ends.
Dialogue: 0,0:18:17.25,0:18:20.48,Default,,0000,0000,0000,,Just to convince you of\Nthat, let's have a look at
Dialogue: 0,0:18:20.48,0:18:21.07,Default,,0000,0000,0000,,another problem.
Dialogue: 0,0:18:22.64,0:18:29.32,Default,,0000,0000,0000,,So we have one 23456\Nand Seven. And let's suppose
Dialogue: 0,0:18:29.32,0:18:35.33,Default,,0000,0000,0000,,this time we pick, say,\Nthe .3 for A.
Dialogue: 0,0:18:36.08,0:18:38.25,Default,,0000,0000,0000,,And the point 74B.
Dialogue: 0,0:18:39.08,0:18:42.63,Default,,0000,0000,0000,,And we're looking at this\Nstraight line. Segments here
Dialogue: 0,0:18:42.63,0:18:46.40,Default,,0000,0000,0000,,between A&B. Again, it's rather\Nobvious from looking at it.
Dialogue: 0,0:18:47.81,0:18:51.17,Default,,0000,0000,0000,,The X value at the\Nmidpoint is 5.
Dialogue: 0,0:18:52.87,0:18:57.11,Default,,0000,0000,0000,,And if we average the X values\Nat the beginning and the end.
Dialogue: 0,0:18:57.90,0:19:00.71,Default,,0000,0000,0000,,Of the line segment\Nthree added to 7.
Dialogue: 0,0:19:02.87,0:19:07.97,Default,,0000,0000,0000,,Divided by two as the average of\Nthe X values is 10 over 2, which
Dialogue: 0,0:19:07.97,0:19:12.86,Default,,0000,0000,0000,,is 5. So in a more general case,\Nin a few minutes time will see
Dialogue: 0,0:19:12.86,0:19:16.13,Default,,0000,0000,0000,,that when we want to find the\Ncoordinates of the midpoint of a
Dialogue: 0,0:19:16.13,0:19:18.89,Default,,0000,0000,0000,,line, we just need to average\Nthe values at either end.
Dialogue: 0,0:19:23.61,0:19:28.36,Default,,0000,0000,0000,,OK, let's look at a more general\None. Suppose we have two points.
Dialogue: 0,0:19:28.36,0:19:31.28,Default,,0000,0000,0000,,Suppose A is the .23, B is the
Dialogue: 0,0:19:31.28,0:19:33.31,Default,,0000,0000,0000,,point. 45
Dialogue: 0,0:19:35.35,0:19:39.58,Default,,0000,0000,0000,,Let's plot those. And then we're\Ngoing to try to find the
Dialogue: 0,0:19:39.58,0:19:42.10,Default,,0000,0000,0000,,coordinates of the midpoint of\Nthe line joining them.
Dialogue: 0,0:19:59.92,0:20:04.10,Default,,0000,0000,0000,,OK. So as the .23,\Nso here's Point A.
Dialogue: 0,0:20:07.16,0:20:10.36,Default,,0000,0000,0000,,And Point B is 45, so be will be
Dialogue: 0,0:20:10.36,0:20:12.84,Default,,0000,0000,0000,,up here. And about that.
Dialogue: 0,0:20:15.40,0:20:16.95,Default,,0000,0000,0000,,And there's my line segment.
Dialogue: 0,0:20:18.89,0:20:19.41,Default,,0000,0000,0000,,Maybe?
Dialogue: 0,0:20:21.18,0:20:24.78,Default,,0000,0000,0000,,We are interested in the\Npoint which is the midpoint
Dialogue: 0,0:20:24.78,0:20:28.02,Default,,0000,0000,0000,,of this line, so let's put\Nthat on there.
Dialogue: 0,0:20:29.40,0:20:30.82,Default,,0000,0000,0000,,And let's call it point P.
Dialogue: 0,0:20:31.68,0:20:34.35,Default,,0000,0000,0000,,And because it's the\Nmidpoint of the line, we
Dialogue: 0,0:20:34.35,0:20:37.62,Default,,0000,0000,0000,,know that this length must\Nbe the same as this length.
Dialogue: 0,0:20:38.68,0:20:41.70,Default,,0000,0000,0000,,AP Is the same length as PB.
Dialogue: 0,0:20:42.93,0:20:44.92,Default,,0000,0000,0000,,Let's put some other\Nlines in here.
Dialogue: 0,0:20:46.33,0:20:47.82,Default,,0000,0000,0000,,A horizontal line through a.
Dialogue: 0,0:20:49.69,0:20:51.05,Default,,0000,0000,0000,,Vertical line through be.
Dialogue: 0,0:20:56.07,0:20:59.47,Default,,0000,0000,0000,,Horizontal line through P and a\Nvertical line through P.
Dialogue: 0,0:21:01.05,0:21:04.31,Default,,0000,0000,0000,,And let's label some of these\Npoints that we've created C.
Dialogue: 0,0:21:05.06,0:21:07.43,Default,,0000,0000,0000,,And D. And he.
Dialogue: 0,0:21:09.74,0:21:13.55,Default,,0000,0000,0000,,Now let's look at the line\Nsegment from A to E.
Dialogue: 0,0:21:15.91,0:21:16.42,Default,,0000,0000,0000,,OK.
Dialogue: 0,0:21:19.44,0:21:20.38,Default,,0000,0000,0000,,Sorry Peter.
Dialogue: 0,0:21:21.89,0:21:22.41,Default,,0000,0000,0000,,Not step.
Dialogue: 0,0:21:23.98,0:21:24.51,Default,,0000,0000,0000,,OK.
Dialogue: 0,0:21:28.01,0:21:28.30,Default,,0000,0000,0000,,Yeah.
Dialogue: 0,0:21:33.22,0:21:35.75,Default,,0000,0000,0000,,Now let's look at the triangle.
Dialogue: 0,0:21:36.39,0:21:40.35,Default,,0000,0000,0000,,ACP. And the
Dialogue: 0,0:21:40.35,0:21:43.07,Default,,0000,0000,0000,,triangle BD. Sorry, the\Ntriangle.
Dialogue: 0,0:21:44.53,0:21:45.01,Default,,0000,0000,0000,,Yeah.
Dialogue: 0,0:21:50.76,0:21:54.38,Default,,0000,0000,0000,,Now let's look at\Nthe triangle ACP.
Dialogue: 0,0:21:55.45,0:21:57.63,Default,,0000,0000,0000,,And the triangle PD.
Dialogue: 0,0:21:58.33,0:21:58.76,Default,,0000,0000,0000,,Be.
Dialogue: 0,0:22:00.48,0:22:03.12,Default,,0000,0000,0000,,Both of these triangles are\Nright angle triangles
Dialogue: 0,0:22:03.12,0:22:05.76,Default,,0000,0000,0000,,because they both have a\Nright angle here.
Dialogue: 0,0:22:08.02,0:22:10.51,Default,,0000,0000,0000,,Both of the triangles have\Nthe same hypotenuse, which
Dialogue: 0,0:22:10.51,0:22:13.28,Default,,0000,0000,0000,,has the same length 'cause\Nthey both have the same
Dialogue: 0,0:22:13.28,0:22:14.39,Default,,0000,0000,0000,,length there in there.
Dialogue: 0,0:22:15.50,0:22:17.28,Default,,0000,0000,0000,,Also, this angle in here.
Dialogue: 0,0:22:18.02,0:22:21.16,Default,,0000,0000,0000,,Is the same as this angle in\Nhere because their corresponding
Dialogue: 0,0:22:21.16,0:22:26.02,Default,,0000,0000,0000,,angles? So in fact, because\Nwe've got two angles and a
Dialogue: 0,0:22:26.02,0:22:29.29,Default,,0000,0000,0000,,side in the first triangle\Ncorresponding to two angles
Dialogue: 0,0:22:29.29,0:22:32.92,Default,,0000,0000,0000,,and the side in the second\Ntriangle, these two triangles
Dialogue: 0,0:22:32.92,0:22:34.73,Default,,0000,0000,0000,,or what we call congruent.
Dialogue: 0,0:22:36.48,0:22:39.49,Default,,0000,0000,0000,,Congruent triangles are\Nidentical in every respect, and
Dialogue: 0,0:22:39.49,0:22:44.00,Default,,0000,0000,0000,,in particular. What this means\Nis that this length from B down
Dialogue: 0,0:22:44.00,0:22:48.21,Default,,0000,0000,0000,,to D. Most corresponding\Nwith the same as this length
Dialogue: 0,0:22:48.21,0:22:49.76,Default,,0000,0000,0000,,from P down to see.
Dialogue: 0,0:22:51.87,0:22:57.37,Default,,0000,0000,0000,,So In other words, because these\Ntwo lengths are the same BD and
Dialogue: 0,0:22:57.37,0:23:00.33,Default,,0000,0000,0000,,PC, which is the same as DE.
Dialogue: 0,0:23:01.48,0:23:03.51,Default,,0000,0000,0000,,Then this point D.
Dialogue: 0,0:23:04.32,0:23:07.54,Default,,0000,0000,0000,,Must be the midpoint of the line\Nfrom A to B.
Dialogue: 0,0:23:09.12,0:23:13.94,Default,,0000,0000,0000,,So the Y coordinate of D\Nmust be the average of
Dialogue: 0,0:23:13.94,0:23:16.13,Default,,0000,0000,0000,,the Y coordinates at B&B.
Dialogue: 0,0:23:17.22,0:23:21.82,Default,,0000,0000,0000,,Now the average of the Y\Ncoordinates at E&B is 3 + 5
Dialogue: 0,0:23:21.82,0:23:26.78,Default,,0000,0000,0000,,/ 2, which is 8 over 2 which\Nis 4. So the Y coordinates
Dialogue: 0,0:23:26.78,0:23:28.55,Default,,0000,0000,0000,,at D must be 4.
Dialogue: 0,0:23:30.32,0:23:31.73,Default,,0000,0000,0000,,What about the X coordinate?
Dialogue: 0,0:23:32.76,0:23:35.09,Default,,0000,0000,0000,,If we look at this line from A
Dialogue: 0,0:23:35.09,0:23:38.76,Default,,0000,0000,0000,,to C. Because the triangles are\Ncongruent, then this distance
Dialogue: 0,0:23:38.76,0:23:43.06,Default,,0000,0000,0000,,from A to see must be the same\Nas the distance from P to D,
Dialogue: 0,0:23:43.06,0:23:46.22,Default,,0000,0000,0000,,which is the same as the\Ndistance from C to E.
Dialogue: 0,0:23:47.08,0:23:50.74,Default,,0000,0000,0000,,So In other words, because these\Ntwo distances are the same, see
Dialogue: 0,0:23:50.74,0:23:52.57,Default,,0000,0000,0000,,must be the midpoint of AE.
Dialogue: 0,0:23:53.52,0:23:55.63,Default,,0000,0000,0000,,And because C is the midpoint of
Dialogue: 0,0:23:55.63,0:24:00.74,Default,,0000,0000,0000,,AE. It's X coordinate must be\Nthe average of the X coordinates
Dialogue: 0,0:24:00.74,0:24:06.88,Default,,0000,0000,0000,,A&E. Now the average of those X\Ncoordinates is 2 + 4, which is 6
Dialogue: 0,0:24:06.88,0:24:11.32,Default,,0000,0000,0000,,/ 2 which is 3. So the X\Ncoordinates here must be 3.
Dialogue: 0,0:24:12.43,0:24:16.85,Default,,0000,0000,0000,,So the X coordinate is 3. We\Nknow the Y coordinate is 3. At
Dialogue: 0,0:24:16.85,0:24:20.96,Default,,0000,0000,0000,,this point we decided that the Y\Ncoordinate was four and we know
Dialogue: 0,0:24:20.96,0:24:22.86,Default,,0000,0000,0000,,the X coordinate is also 4.
Dialogue: 0,0:24:24.05,0:24:28.87,Default,,0000,0000,0000,,So now we can read off what the\Ncoordinates of point PRP will
Dialogue: 0,0:24:28.87,0:24:30.73,Default,,0000,0000,0000,,have an X Coordinate A3.
Dialogue: 0,0:24:32.79,0:24:34.85,Default,,0000,0000,0000,,And AY coordinate of four.
Dialogue: 0,0:24:37.42,0:24:39.75,Default,,0000,0000,0000,,Now that's very complicated. It\Nseems very complicated, but in
Dialogue: 0,0:24:39.75,0:24:42.08,Default,,0000,0000,0000,,practice it's very simple. Let's\Njust look at these numbers
Dialogue: 0,0:24:42.08,0:24:45.03,Default,,0000,0000,0000,,again. The X coordinate at the
Dialogue: 0,0:24:45.03,0:24:50.39,Default,,0000,0000,0000,,midpoint. Which is 3 is just the\Naverage of the two X coordinates
Dialogue: 0,0:24:50.39,0:24:55.95,Default,,0000,0000,0000,,2 + 46 / 2 is 3, the Y\Ncoordinate at the mid .4 is just
Dialogue: 0,0:24:55.95,0:24:59.42,Default,,0000,0000,0000,,the average of the two Y\Ncoordinates. Five and three,
Dialogue: 0,0:24:59.42,0:25:01.84,Default,,0000,0000,0000,,which is 8 / 2 is 4.
Dialogue: 0,0:25:02.64,0:25:09.55,Default,,0000,0000,0000,,So in general, if we have\Na situation where we've got a
Dialogue: 0,0:25:09.55,0:25:13.01,Default,,0000,0000,0000,,point AX1Y one and point BX2Y2.
Dialogue: 0,0:25:13.85,0:25:18.47,Default,,0000,0000,0000,,The coordinates of the midpoint\Nof the line AV are the average
Dialogue: 0,0:25:18.47,0:25:20.01,Default,,0000,0000,0000,,of the X coordinates.
Dialogue: 0,0:25:23.48,0:25:25.42,Default,,0000,0000,0000,,And the average of the Y\Ncoordinates.
Dialogue: 0,0:25:29.49,0:25:31.60,Default,,0000,0000,0000,,Let's just look at a\Nsimple example just to
Dialogue: 0,0:25:31.60,0:25:32.30,Default,,0000,0000,0000,,hammer that home.
Dialogue: 0,0:25:34.46,0:25:40.90,Default,,0000,0000,0000,,Suppose we have a point A with\Ncoordinates 2 - 4 and be with
Dialogue: 0,0:25:40.90,0:25:45.50,Default,,0000,0000,0000,,coordinates minus four and\Nthree. So two points and were
Dialogue: 0,0:25:45.50,0:25:50.56,Default,,0000,0000,0000,,interested in the coordinates of\Nthe midpoint of the line AB.
Dialogue: 0,0:25:51.84,0:25:53.31,Default,,0000,0000,0000,,Select this point BP.
Dialogue: 0,0:25:54.81,0:25:59.13,Default,,0000,0000,0000,,The X coordinate will be the\Naverage of the two X
Dialogue: 0,0:25:59.13,0:26:03.06,Default,,0000,0000,0000,,coordinates, which is 2. Added 2\N- 4 over 2.
Dialogue: 0,0:26:04.67,0:26:08.35,Default,,0000,0000,0000,,And the Y coordinate will be the\Naverage of the two Y
Dialogue: 0,0:26:08.35,0:26:11.42,Default,,0000,0000,0000,,coordinates, which is minus four\Nadded to 3 / 2.
Dialogue: 0,0:26:12.13,0:26:16.39,Default,,0000,0000,0000,,And if we tie do that up, will\Nget 2 subtract 4, which is minus
Dialogue: 0,0:26:16.39,0:26:18.38,Default,,0000,0000,0000,,2 over 2, which is minus one.
Dialogue: 0,0:26:19.25,0:26:24.83,Default,,0000,0000,0000,,And minus 4 + 3 is minus 1 - 1\Nover 2 is minus 1/2.
Dialogue: 0,0:26:26.03,0:26:30.61,Default,,0000,0000,0000,,So there is the coordinates\Nof the midpoint of the line
Dialogue: 0,0:26:30.61,0:26:31.44,Default,,0000,0000,0000,,joining amb.