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Distance Formula - ConceptFREE
Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school
Brian was a geometry teacher through the Teach for America program and started the geometry program at his school
Using what we know about the Pythagorean theorem, we are able to derive the distance formula which is used to find the straight distance between two points in a coordinate plane. The distance formula is a standard formula that allows us to plug a set of coordinates into the formula and easily calculate the distance between the two.
If you have two points, let's call them
A and B, somewhere in a coordinate
plane, and we call A X1 and Y1. That's
the ordered pair A and we say B has ordered
pair X2 and Y2. We can calculate the
direct straight line distance between them.
using what we know about the Pythagorean theorem.
You might say Mr. Mccall how are we going
to use a Pythagorean on a line that's
diagonal like that, you don't even have a triangle.
Well, what I'm going to do, I'm going to
draw in one leg of that triangle that's
going to be parallel to my X axis.
And we're going to draw in another leg
of that triangle which is parallel to
the Y axis.
I know the X axis and Y axis are perpendicular
to each other which means that
this must be a right triangle.
If we want to find out the distance between
A and B, first we need to say, well,
what are the lengths of my legs.
The reason why that's important is because
we're going to use A squared plus
B squared equals C squared.
So A is going to be one of my legs.
And let's call it the leg that's
parallel to the X axis.
Well, this point right here is going to
be the point not X 1, but it's going
to be X2 and Y1.
Because notice the only thing that's changed
from A to this corner is my value
of X. If these two lines are parallel,
then Y1 will stay the same.
So if I want to find the distance between
these two, all I need to do is subtract
So this distance is X2 minus X1. That
difference will tell me how far away
those points are.
So I'm going to say that A is X2 minus X1.
If I find B, B is going to be the
other leg of this triangle.
So just like I said that this horizontal
distance was the difference of our
axis, the vertical distance will be
the vertical distance of our Yes.
So this will be Y2 minus Y1.
So B is going to equal Y2 minus Y1.
And the hypotenuse C we could
say is D, our distance.
Or I guess if you want to, you could
say that this is line segment AB.
Either way, you're trying to
find your hypotenuse here.
So let's substitute in what we know.
Well, we said -- if I use a different marker
-- we said we were going to use
the Pythagorean theorem, and A is X2 minus X1.
I'm going to say we're going
to have X2 minus X1 squared.
So all I'm doing is substituting in here.
B we said was Y2 minus Y1, starting
to add Y2 minus Y1 squared.
And C we said is our distance, AB.
And that's going to be squared.
So if you want to know the square of the
distance, in your coordinate
plane, you're going to subtract
your Xs square them. Subtract your Y square
them and add them up.
Well, that's not quite useful.
So we're going to take the square root
of both sides, because the square of
a distance doesn't
help me that much.
So I'm going to say that the square root
of X2 minus X1 squared plus Y2
minus Y1 squared is equal
to this distance AB.
And, voila, we have our distance formula.
So the distance between any two points
in space is going to be the difference
of your Xs squared plus the difference
of your Ys squared.
Now, some of you might be thinking,
Mr. McCall, I know that the square root of something
squared is whatever that base term is.
Now, you cannot say that either of these
squares are going to come out.
The reason is we have this expression by this plus sign.
So if this whole thing was being squared,
then, yes, something could come out
of this square root.
But since we have this plus sign it's
going to stay the way this is.
So the keys to using this formula are
subtracting your Xs, subtracting your
Ys, squaring those and then
taking the square root.
We got this formula by using
the Pythagorean theorem.
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