Question

If A(-2,10) and B(4,-2), then AB= ?

Answer 1

13.42

Explanation:

Hi,

This is high school match, so I'm assuming you know the Pythagorean theorem. They may want you to use "the distance formula," but this is the same thing, but it shows it in a way that lets you understand what's going on.

Refer to the drawing I attached. That's from an online graphing utility I found, and I don't know why it kind of cut off Point A, but that's it at the top at (-2,10).

So that's the 2 points plotted, A & B, connected by the green line. The easiest way to do these kinds of problems is to let that be the hypotenuse of a right triangle, then draw in (or imagine) the other two legs.

I've done that in blue by adding point C at (-2,-2). Do those -2's look familiar? They should, because they're the x coordinate in A and the y coordinate in B.

But you don't have to know any of that: on your graph paper just draw a horizontal line left from B and down from A. Where they intersect they should make a right triangle. Call that point C or whatever you want.

Now how long is each of those legs?

• AC is 12 units long (count 'em if you want, or add 10 above the x-axis to 2 below it).
• BC is 6 units long.

Now we have a right triangle and we know the lengths of 2 of its sides. Knowing those 3 things (right angle, length of 2 sides), we can figure out the length of the 3rd side, the hypotenuse, which is the distance we're looking for. (We can also find the other 2 angles, btw.)

A really smart Greek mathematician named Pythagoras figured that out about 2,500 years ago (and the Egyptians had gotten close 1,300 years before that).

What did he tell us? Simply that
`a^(2) + b^(2) = c^(2)`

Looks simple, right? But it's powerful, so memorize it.

What it says is that if you have a right triangle, take the lengths of the two shorter legs, square them and sum them, and that sum will be the square of the length of the long side, the hypotenuse.

a and b are the line segments AC and BC, and you can call either one b or a.

c is the hypotenuse, and that's the unknown we want to solve for, so let's do that. Turn the equation around first:

`c^(2) = a^(2) + b^(2)`

Do you know how to "get c by itself"? Since it's being squared, you need to take the square root of it, right? The square root cancels out the exponent of 2, leaving just c.

But always remember that whatever you do to one side of an equation (an equality), you have to do to the other side. Otherwise the two sides no longer equal each other.

So we have to take the square root of
`(a^(2) + b^(2))`. so just throw a long square root symbol over that whole phrase, to get this:

`c = \sqrt{(a^(2) + b^(2))`

That's the basis of "the distance formula," but a is (X₁ - X₂) and b is (Y₁ - Y₂). But look at the graph: that's exactly how we determined the lengths of the horizontal and vertical legs. Be careful with the signs, but it works.

Anyway, now that you know HOW to do it, what are a and b in this problem? 12 and 6, right? Or 6 and 12, doesn't matter.

• 12 squared is 144
• 6 squared is 36
• Add them to get 180.
• Take the square root of that to get 13.42.
• That's c, which is the answer.

If you like equations better, it looks like this:

c =
`\sqrt{12^(2) + 6^(2) } = √(144 + 36) = √(180) = 13.42`

We should always check our answers though:

• Does it kind of make sense? Well yeah, because in a right triangle the hypotenuse always has to be the longest leg. And 13.42 is longer than 12, and of course longer than 6.
• We can plug the numbers back into the theorem and see if it works out:
  `13.42^(2) = 12^(2) + 6^(2)` --> 180 = 144 + 36 = 180 So yes, it checks out.

So there you go, now you know how to do any of these types of problems they throw at you.

Answer 2

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{10})\qquad B(\stackrel{x_2}{4}~,~\stackrel{y_2}{-2})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ AB=√((~~4 - (-2)~~)^2 + (~~-2 - 10~~)^2)\implies AB=√((4 +2)^2 + (-2 -10)^2) \\\\\\ AB=√( (6)^2 + (-12)^2) \implies AB=√( 180 )\implies AB\approx 13.42`