Let's look at rectangular coordinates.
To discuss this topic, we need to draw the rectangular coordinate system,
which is composed of two lines that are drawn perpendicular
and intersect at a given point.
The horizontal line is called the x-axis and the vertical line is called the y-axis.
The points in the plane that the two axes divide up the plane into,
those regions are called the quadrants.
And they are numbered counterclockwise one, two, three, four.
If you study trig that will be very important.
Now we can locate points with their coordinates.
And each point in the plane is represented by a set of two coordinates,
an x-coordinate and y-coordinate.
So let’s just pick a point.
We are going to pick a point there, in quadrant one.
From the origin, which has coordinates (0,0),
we move over on the x-axis however many units x is,
and then on the y-axis we move over y units and we locate that point.
We might take an example.
Let's plot the point (-1,4). Let’s see if you remember how to do that.
We always start at the origin,
and from there I'm going to move negative 1,
or in the left direction on the x-axis one unit.
And then from there I'm going to move 4 units up,
and I plot the point (-1,4).
We need this coordinate system to locate points,
eventually to graph, but in this material
we want to look at distance between two points
and the midpoint of a segment.
We will need two formulas for this.
If we have two points in the coordinate system such as
point P with coordinates (x1,y1) and point Q with coordinates (x2,y2),
we can find the distance between the two points.
That formula is the square root of the difference of the x-coordinates squared
plus the difference of the y-coordinates squared.
The other formula the we are going to use is to find the coordinates of the midpoint.
If we were to draw a segment between points P and Q,
we would locate the midpoint exactly halfway between the two points.
Because that point is exactly halfway between,
we find the coordinates by taking the average of the x-coordinates,
and then the average of the y-coordinates.
We will work two examples using these formulas.
The main thing you want to remember about these two formulas:
the distance formula is the difference of the coordinates,
the midpoint formula we are taking the sum.
Many students want to take the difference on both of the formulas.
So we will look at an example now using some numbers.
Example one. We are given two points P and Q.
P has coordinates (-2,3) and Q has coordinates (6,-5).
There are two thing we want to find:
the distance between the two points and,
the midpoint of the segment connecting points P and Q.
To work this problem, I am going to first draw our rectangular coordinate system,
and I am going to plot the two points.
So we will get a visual picture of what this looks like.
(-2,3), that point will be located in quadrant two,
moving 2 units to the left, 3 up.
Then point Q, we will go 6 units to the right on the x-axis,
and 5 units down and locate point Q.
So we see where our two points are.
But all we want to do is find the distance.
So we're going to take the square root of the difference of the x-coordinates,
and it does not matter if this point P is the coordinates (x1,y1)
and these are (x2,y2) or the reverse.
So I am just going to take negative 2 and subtract 6 then we'll square that.
We will add to that 3 minus negative 5 squared.
I am going to keep the order the same.
Now in order to simplify this we must follow order of operations.
So we are going to combine negative 2 and negative 6, which is negative 8.
Squared is a positive 64.
3 minus a negative 5 is a positive 8.
Squared is a positive 64.
Now you might be tempted to take the square root of 64, which is 8,
and add it to the square root of 64, which is 8.
So you would think that the distance between the two points is 16,
but that is incorrect.
Remember again, by order of operations we're going to add 64 plus 64,
and we have the square root of 128.
This is an exact answer but it's not simplified.
So let's simplify it. In order to do that, I want to break 128 into two factors,
one of which is a perfect square. That is of course 64.
And then take the square root of 64, which is 8.
So 8 times the square root of 2 is our final answer.
That represents the distance between P and Q.
Now for the b part of this problem,
we are going to use our midpoint formula to find the coordinates of the midpoint.
So this time we're not subtracting, we are adding.
A negative 2 plus 6 divided by 2. And then positive 3 with a negative 5 divided by 2.
We will do the arithmetic.
Negative 2 plus 6 is 4. Divided by 2 is positive 2.
3 and a negative 5 is -2. Divided by positive 2 is a negative 1.
And of course we use parentheses to enclose the x and the y-coordinates.
This is our standard notation for the coordinates of a point.
If we plot that point it would be here: 2 to the right and 1 down.
And we can see that it does look like the midpoint
if we connected the two points with a segment.
So this example shows us basically how to work
with the distance formula and the midpoint formula.
We will look at a second example using the distance formula.
Example two.
Find all points having a y coordinate of negative 3
whose distance from the point P, that has coordinates (1,2), is 13.
To work this problem I want to first draw a coordinate system,
and plot the point P. Notice it's in quadrant one.
I will put the coordinated by point P.
And then let's see what we are looking for.
We want to find all points, we don't know how many that is,
that have a y-coordinate of negative 3.
The y-coordinate is negative 3.
Notice it did not say anything about the x-coordinates.
That is what we have to find. So where is y equals negative 3?
Well if we go down 3 units on the y-axis let's just draw dotted line,
and think about where those points could be.
There are points all along this dotted line that could be one of the points,
if we have more than one, that would answer this problem.
So let's just arbitrarily pick one. Let's call it (x,-3)
since we're going to be solving for x.
So the other part of the problem is that
the distance between point P and each of these points is 13.
So from P to this point, I know that the distance is 13.
So 13 is going to equal the square root of the difference of the x-coordinates.
So let’s do x minus 1 squared, plus then the difference of the y-coordinates squared.
So I'm going to have negative 3 minus 2, squared.
Okay, because we are solving for x and not for distance,
we are going to have to work this a little differently.
But, I am going to just simplify this part first.
Then we will see what we can do with this.
Negative 3 minus 2 is negative 5. Squared is 25.
Now remember I cannot take the square root of the x minus 1 squared
and the 25 separately.
So I'm going to have to do something to remove that radical.
Well I know I can square both sides of the equation.
13 squared is 169. And the square root of this whole quantity squared,
will leave us with x minus 1 squared plus 25.
So this is different from what we did in the first example.
We are solving an equation.
Now we can subtract the 25 from both sides.
We end up with 144 equals the square of x minus 1.
Now at this point there are two approaches:
we can use the square root method, take the square root of both sides,
or we can actually F.O.I.L. this out, collect all the like terms,
set it equal to zero, factor, and work it that way.
The choice is yours. I am going to choose to use the square root method.
Now to square root, I'm going to need the square root,
the positive and negative square root, of 144.
So that's going to give me plus or minus 12 equals x minus 1.
Because the square root x minus 1 squared, will give me x minus 1.
Now we have plus or minus 12.
So we are going to have 12 equals x minus 1,
or negative 12 equals x minus 1.
Solving for x by adding 1, we get x equals 13 and x is negative 11.
So notice, we have two points.
I am going to write, because we want points,
I am going to write the points with the coordinates as an ordered pair of numbers.
So one answer will be the point (13,-3).
The second answer will be the point with the coordinates (-11,-3).
So this gives us a second example.
This problem is a little more difficult than example one,
and at this point, lead for you to practice working
all of these homework problems until you feel comfortable
in simplifying and finding distance and midpoint given any two points.