_{2}-x

_{1})

^{2 }+ (y

_{2}-y

_{1})

^{2}

^{2 }+ (-7-(-7))

^{2}

^{2 }+ (0)

^{2}

need help finding the distance between those to points for geometry

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Hey Joey, you wouldn't even have to use the Pythagorean formula, all you need to know and use is the distance formula which is:

d= √(x_{2}-x_{1})^{2 }+ (y_{2}-y_{1})^{2}

so √(6-1)^{2 }+ (-7-(-7))^{2}

√(5)^{2 }+ (0)^{2}

√25

d=5

so the distance is 5

Hey Joey -

Have you run across the Pythagorean Theorem yet? It's probably the most useful equation you'll take away from Geometry class - you'll use it everywhere. Fortunately it's pretty simple to remember.

A^{2} + B^{2} = C^{2}

The Wikipedia article on the Pythagorean Theorem has a couple of good images to help show what this means, but in a nutshell:

Take a triangle. One that has a right angle. A right triangle.

"A" stands for the length of one side.

"B" stands for the length of another side.

"C" stands for the length of the third side - the one opposite the right angle.

Why am I talking about triangles when you only have two points? In your question, those two points are on a flat line - they have the same y-value (they're both seven units below the x-axis). But that may not be the case in your next problem. Fortunately, that doesn't keep us from solving your question here.

Let's make A represent the distance along the x-axis between your points (between 1 and 6). There are five units between those points (6 - 1 = 5, or 1 - 6= -5, either way it's five units forward or back from one to the other). So A is 5.

Let's have B represent the distance along the y-axis between your points (-7 and -7). They're both at the same height (have the same y-value), so there's no distance between them in that direction (-7 - -7 = -7 + 7 = 0). So B is zero.

Once you've got two of the three variables (A and B), you can find out what C is. Using the Pythagorean Theorem shown above (A^{2} + B^{2} = C^{2}), fill in the values you know:

5^{2} + 0^{2} = C^{2}

You can then solve the equation (see the steps below), but in this special case, you can also simplify it first if you know that 0^{2} is 0, which adds nothing to the value and can be safely left out of the matter. Rewriting without the zero, you get

5^{2} = C^{2}

Take a look at it - if both sides of the equation are squared, then removing that exponent from both sides is ok, and leaves us with

5 = C

Which means that the length of the third leg of the triangle (the distance between those two points of yours) is 5.

The long way through that equation:

5^{2} is 5 x 5 is 25

0^{2} is 0 x 0 is 0

so

25 + 0 = C^{2}

so

C = √25 (the square root of 25), which is 5.

This method works when the two points are not at the same level (have the same y-value) just as well.

Let's say that instead of points at (1,-7) and (6,-7), you have points at (10, 3) and (7, -1). Run those numbers through the Theorem....

.... no peeking....

Ok - here's what you probably got:

starting with

A^{2} + B^{2} = C^{2}

you filled in what you know:

A = 10 - 7 = 3

B = 3 - -1 = 3 + 1 = 4

so

3^{2} + 4^{2} = C^{2}

3^{2} is 3 x 3, which is 9

4^{2} is 4 x 4, which is 16

so

9 + 16 = C^{2}

so

25 = C^{2}

so

C = √25 (the square root of 25), which is 5.

C = √25 (the square root of 25), which is 5.

No, the answer won't always be 5. It was here because I set it up that way - there's a special relationship between 3, 4, and 5 that is unique and shows up a LOT (3^{2} + 4^{2} = 5^{2}). Keep an eye out and you'll see that pattern everywhere - in buildings, in art, and on math tests.

For an answer that's not 5, try these points:

(1,1) and (1,3)

(6,-3) and (-2,-3)

(2,2) and (0,0) (hint: this last answer you'll have to just leave as the square root of something - which you'll do a lot in future classes)

Hope this helps.

-sj

Hi Joey;

The y-coordinate is the same for both coordinates; -7.

Therefore, the equation we need is...

x-x_{1}

1-6=-5

The result should be the absolute value of -5. This is because it is the distance whether we measure from the first point to the second, or the second point to the first.

The distance is...

5 units

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