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# Distance Formula - Concept

FREE###### Brian McCall

###### Brian McCall

**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

my Xs.

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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###### Brian McCall

B.S. in Chemical Engineering, University of Wisconsin

J.D. University of Wisconsin Law School (magna cum laude)

He doesn't beat around the bush. His straightforward teaching style is effective and his subtle midwestern accent is engaging. There's never a dull moment with him.

so my teacher can't explain this in 5 weeks but I learn this in less than 3 minutes”

its hard to focus when the teacher is really really really goodlooking”

i like how it took you 3 minutes and 8 seconds to accomplish what my teacher couldn't in 3 days”

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