WEBVTT
00:00:06.700 --> 00:00:09.600
Let's look at rectangular coordinates.
00:00:10.950 --> 00:00:16.650
To discuss this topic, we need to draw the rectangular coordinate system,
00:00:16.650 --> 00:00:23.450
which is composed of two lines that are drawn perpendicular
00:00:23.450 --> 00:00:29.750
and intersect at a given point.
00:00:29.750 --> 00:00:39.150
The horizontal line is called the x-axis and the vertical line is called the y-axis.
00:00:39.150 --> 00:00:46.750
The points in the plane that the two axes divide up the plane into,
00:00:46.750 --> 00:00:50.250
those regions are called the quadrants.
00:00:51.850 --> 00:01:01.250
And they are numbered counterclockwise one, two, three, four.
00:01:01.250 --> 00:01:05.850
If you study trig that will be very important.
00:01:05.850 --> 00:01:11.750
Now we can locate points with their coordinates.
00:01:14.750 --> 00:01:21.850
And each point in the plane is represented by a set of two coordinates,
00:01:21.850 --> 00:01:24.150
an x-coordinate and y-coordinate.
00:01:24.150 --> 00:01:25.650
So let’s just pick a point.
00:01:25.650 --> 00:01:28.950
We are going to pick a point there, in quadrant one.
00:01:30.000 --> 00:01:35.250
From the origin, which has coordinates (0,0),
00:01:37.350 --> 00:01:44.000
we move over on the x-axis however many units x is,
00:01:44.950 --> 00:01:56.350
and then on the y-axis we move over y units and we locate that point.
00:01:56.350 --> 00:01:59.500
We might take an example.
00:01:59.500 --> 00:02:06.650
Let's plot the point (-1,4). Let’s see if you remember how to do that.
00:02:06.650 --> 00:02:09.150
We always start at the origin,
00:02:09.150 --> 00:02:14.000
and from there I'm going to move negative 1,
00:02:14.000 --> 00:02:19.000
or in the left direction on the x-axis one unit.
00:02:19.000 --> 00:02:26.150
And then from there I'm going to move 4 units up,
00:02:26.150 --> 00:02:30.350
and I plot the point (-1,4).
00:02:30.350 --> 00:02:35.500
We need this coordinate system to locate points,
00:02:35.500 --> 00:02:39.450
eventually to graph, but in this material
00:02:39.450 --> 00:02:43.000
we want to look at distance between two points
00:02:43.000 --> 00:02:46.450
and the midpoint of a segment.
00:02:46.450 --> 00:02:51.650
We will need two formulas for this.
00:02:55.500 --> 00:02:59.850
If we have two points in the coordinate system such as
00:02:59.850 --> 00:03:17.350
point P with coordinates (x1,y1) and point Q with coordinates (x2,y2),
00:03:17.350 --> 00:03:21.650
we can find the distance between the two points.
00:03:21.650 --> 00:03:37.250
That formula is the square root of the difference of the x-coordinates squared
00:03:37.250 --> 00:03:46.350
plus the difference of the y-coordinates squared.
00:03:48.150 --> 00:03:57.600
The other formula the we are going to use is to find the coordinates of the midpoint.
00:03:57.600 --> 00:04:03.750
If we were to draw a segment between points P and Q,
00:04:03.750 --> 00:04:10.850
we would locate the midpoint exactly halfway between the two points.
00:04:13.000 --> 00:04:18.750
Because that point is exactly halfway between,
00:04:18.750 --> 00:04:27.650
we find the coordinates by taking the average of the x-coordinates,
00:04:27.650 --> 00:04:32.000
and then the average of the y-coordinates.
00:04:33.250 --> 00:04:37.950
We will work two examples using these formulas.
00:04:37.950 --> 00:04:42.000
The main thing you want to remember about these two formulas:
00:04:42.000 --> 00:04:46.350
the distance formula is the difference of the coordinates,
00:04:46.350 --> 00:04:50.500
the midpoint formula we are taking the sum.
00:04:50.500 --> 00:04:55.850
Many students want to take the difference on both of the formulas.
00:04:55.850 --> 00:05:00.350
So we will look at an example now using some numbers.
00:05:04.250 --> 00:05:07.850
Example one. We are given two points P and Q.
00:05:07.850 --> 00:05:13.650
P has coordinates (-2,3) and Q has coordinates (6,-5).
00:05:13.650 --> 00:05:15.950
There are two thing we want to find:
00:05:15.950 --> 00:05:18.350
the distance between the two points and,
00:05:18.350 --> 00:05:22.450
the midpoint of the segment connecting points P and Q.
00:05:24.350 --> 00:05:30.500
To work this problem, I am going to first draw our rectangular coordinate system,
00:05:30.500 --> 00:05:34.650
and I am going to plot the two points.
00:05:34.650 --> 00:05:37.850
So we will get a visual picture of what this looks like.
00:05:37.850 --> 00:05:44.350
(-2,3), that point will be located in quadrant two,
00:05:44.350 --> 00:05:48.250
moving 2 units to the left, 3 up.
00:05:48.250 --> 00:05:57.850
Then point Q, we will go 6 units to the right on the x-axis,
00:05:57.850 --> 00:06:03.350
and 5 units down and locate point Q.
00:06:03.350 --> 00:06:06.450
So we see where our two points are.
00:06:07.650 --> 00:06:11.250
But all we want to do is find the distance.
00:06:13.850 --> 00:06:20.000
So we're going to take the square root of the difference of the x-coordinates,
00:06:20.000 --> 00:06:25.350
and it does not matter if this point P is the coordinates (x1,y1)
00:06:25.350 --> 00:06:29.500
and these are (x2,y2) or the reverse.
00:06:29.500 --> 00:06:37.900
So I am just going to take negative 2 and subtract 6 then we'll square that.
00:06:37.900 --> 00:06:44.950
We will add to that 3 minus negative 5 squared.
00:06:44.950 --> 00:06:47.100
I am going to keep the order the same.
00:06:47.100 --> 00:06:53.000
Now in order to simplify this we must follow order of operations.
00:06:53.000 --> 00:06:59.700
So we are going to combine negative 2 and negative 6, which is negative 8.
00:06:59.700 --> 00:07:03.000
Squared is a positive 64.
00:07:03.000 --> 00:07:07.350
3 minus a negative 5 is a positive 8.
00:07:07.350 --> 00:07:11.150
Squared is a positive 64.
00:07:11.150 --> 00:07:16.250
Now you might be tempted to take the square root of 64, which is 8,
00:07:16.250 --> 00:07:19.500
and add it to the square root of 64, which is 8.
00:07:19.500 --> 00:07:23.450
So you would think that the distance between the two points is 16,
00:07:23.450 --> 00:07:25.000
but that is incorrect.
00:07:25.000 --> 00:07:32.650
Remember again, by order of operations we're going to add 64 plus 64,
00:07:32.650 --> 00:07:36.950
and we have the square root of 128.
00:07:36.950 --> 00:07:42.000
This is an exact answer but it's not simplified.
00:07:42.000 --> 00:07:50.350
So let's simplify it. In order to do that, I want to break 128 into two factors,
00:07:50.350 --> 00:07:56.400
one of which is a perfect square. That is of course 64.
00:07:56.400 --> 00:08:02.150
And then take the square root of 64, which is 8.
00:08:02.150 --> 00:08:06.750
So 8 times the square root of 2 is our final answer.
00:08:06.750 --> 00:08:10.450
That represents the distance between P and Q.
00:08:12.000 --> 00:08:13.850
Now for the b part of this problem,
00:08:13.850 --> 00:08:22.750
we are going to use our midpoint formula to find the coordinates of the midpoint.
00:08:22.750 --> 00:08:26.500
So this time we're not subtracting, we are adding.
00:08:26.500 --> 00:08:42.150
A negative 2 plus 6 divided by 2. And then positive 3 with a negative 5 divided by 2.
00:08:42.150 --> 00:08:45.100
We will do the arithmetic.
00:08:45.100 --> 00:08:52.450
Negative 2 plus 6 is 4. Divided by 2 is positive 2.
00:08:52.450 --> 00:09:01.150
3 and a negative 5 is -2. Divided by positive 2 is a negative 1.
00:09:01.150 --> 00:09:08.150
And of course we use parentheses to enclose the x and the y-coordinates.
00:09:08.150 --> 00:09:12.750
This is our standard notation for the coordinates of a point.
00:09:12.750 --> 00:09:19.950
If we plot that point it would be here: 2 to the right and 1 down.
00:09:19.950 --> 00:09:23.350
And we can see that it does look like the midpoint
00:09:23.350 --> 00:09:29.450
if we connected the two points with a segment.
00:09:29.450 --> 00:09:34.500
So this example shows us basically how to work
00:09:34.500 --> 00:09:38.450
with the distance formula and the midpoint formula.
00:09:38.450 --> 00:09:42.750
We will look at a second example using the distance formula.
00:09:45.350 --> 00:09:46.450
Example two.
00:09:47.150 --> 00:09:51.450
Find all points having a y coordinate of negative 3
00:09:51.450 --> 00:09:58.850
whose distance from the point P, that has coordinates (1,2), is 13.
00:10:00.000 --> 00:10:04.950
To work this problem I want to first draw a coordinate system,
00:10:04.950 --> 00:10:11.350
and plot the point P. Notice it's in quadrant one.
00:10:11.350 --> 00:10:15.250
I will put the coordinated by point P.
00:10:15.250 --> 00:10:18.350
And then let's see what we are looking for.
00:10:18.350 --> 00:10:22.900
We want to find all points, we don't know how many that is,
00:10:22.900 --> 00:10:26.850
that have a y-coordinate of negative 3.
00:10:26.850 --> 00:10:29.450
The y-coordinate is negative 3.
00:10:29.450 --> 00:10:33.350
Notice it did not say anything about the x-coordinates.
00:10:33.350 --> 00:10:40.350
That is what we have to find. So where is y equals negative 3?
00:10:40.350 --> 00:10:46.850
Well if we go down 3 units on the y-axis let's just draw dotted line,
00:10:46.850 --> 00:10:49.950
and think about where those points could be.
00:10:49.950 --> 00:10:57.600
There are points all along this dotted line that could be one of the points,
00:10:57.600 --> 00:11:01.900
if we have more than one, that would answer this problem.
00:11:01.900 --> 00:11:10.100
So let's just arbitrarily pick one. Let's call it (x,-3)
00:11:10.100 --> 00:11:13.000
since we're going to be solving for x.
00:11:13.000 --> 00:11:16.500
So the other part of the problem is that
00:11:16.500 --> 00:11:22.650
the distance between point P and each of these points is 13.
00:11:22.650 --> 00:11:29.500
So from P to this point, I know that the distance is 13.
00:11:29.500 --> 00:11:37.350
So 13 is going to equal the square root of the difference of the x-coordinates.
00:11:37.350 --> 00:11:46.850
So let’s do x minus 1 squared, plus then the difference of the y-coordinates squared.
00:11:46.850 --> 00:11:54.150
So I'm going to have negative 3 minus 2, squared.
00:11:55.650 --> 00:12:02.250
Okay, because we are solving for x and not for distance,
00:12:02.250 --> 00:12:05.000
we are going to have to work this a little differently.
00:12:05.000 --> 00:12:08.350
But, I am going to just simplify this part first.
00:12:10.000 --> 00:12:12.450
Then we will see what we can do with this.
00:12:13.900 --> 00:12:20.150
Negative 3 minus 2 is negative 5. Squared is 25.
00:12:20.150 --> 00:12:26.600
Now remember I cannot take the square root of the x minus 1 squared
00:12:26.600 --> 00:12:29.650
and the 25 separately.
00:12:29.650 --> 00:12:34.150
So I'm going to have to do something to remove that radical.
00:12:34.150 --> 00:12:43.150
Well I know I can square both sides of the equation.
00:12:43.150 --> 00:12:52.600
13 squared is 169. And the square root of this whole quantity squared,
00:12:52.600 --> 00:12:59.600
will leave us with x minus 1 squared plus 25.
00:12:59.600 --> 00:13:03.850
So this is different from what we did in the first example.
00:13:03.850 --> 00:13:05.750
We are solving an equation.
00:13:05.750 --> 00:13:12.350
Now we can subtract the 25 from both sides.
00:13:13.450 --> 00:13:18.950
We end up with 144 equals the square of x minus 1.
00:13:18.950 --> 00:13:22.950
Now at this point there are two approaches:
00:13:22.950 --> 00:13:27.850
we can use the square root method, take the square root of both sides,
00:13:27.850 --> 00:13:32.750
or we can actually F.O.I.L. this out, collect all the like terms,
00:13:32.750 --> 00:13:36.800
set it equal to zero, factor, and work it that way.
00:13:36.800 --> 00:13:41.150
The choice is yours. I am going to choose to use the square root method.
00:13:41.150 --> 00:13:45.000
Now to square root, I'm going to need the square root,
00:13:45.000 --> 00:13:54.000
the positive and negative square root, of 144.
00:13:54.000 --> 00:14:01.450
So that's going to give me plus or minus 12 equals x minus 1.
00:14:01.450 --> 00:14:08.100
Because the square root x minus 1 squared, will give me x minus 1.
00:14:08.100 --> 00:14:11.500
Now we have plus or minus 12.
00:14:11.500 --> 00:14:16.450
So we are going to have 12 equals x minus 1,
00:14:16.450 --> 00:14:22.350
or negative 12 equals x minus 1.
00:14:22.350 --> 00:14:32.450
Solving for x by adding 1, we get x equals 13 and x is negative 11.
00:14:32.450 --> 00:14:35.350
So notice, we have two points.
00:14:35.350 --> 00:14:38.900
I am going to write, because we want points,
00:14:38.900 --> 00:14:43.850
I am going to write the points with the coordinates as an ordered pair of numbers.
00:14:43.850 --> 00:14:51.150
So one answer will be the point (13,-3).
00:14:51.900 --> 00:15:00.850
The second answer will be the point with the coordinates (-11,-3).
00:15:01.8500 --> 00:15:04.500
So this gives us a second example.
00:15:04.500 --> 00:15:08.350
This problem is a little more difficult than example one,
00:15:08.350 --> 00:15:12.900
and at this point, lead for you to practice working
00:15:12.900 --> 00:15:15.500
all of these homework problems until you feel comfortable
00:15:15.500 --> 00:15:23.200
in simplifying and finding distance and midpoint given any two points.