whats the distance between the points (1,-7) and (6,-7)

Question

need help finding the distance between those to points for geometry 

Answer 1

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= √(x₂-x₁)² + (y₂-y₁)²

 

so √(6-1)² + (-7-(-7))²

 

√(5)² + (0)²

 

√25

 

d=5

 

so the distance is 5

Answer 2

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.

 

A² + B² = C²

 

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 (A² + B² = C²), fill in the values you know:

 

5² + 0² = C²

 

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 0² is 0, which adds nothing to the value and can be safely left out of the matter.  Rewriting without the zero, you get 

 

5² = C²

 

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:

5² is 5 x 5 is 25

0² is 0 x 0 is 0

  so 

25 + 0 = C²

 

  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

A² + B² = C²

  you filled in what you know:

A = 10 - 7 = 3

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

  so

3² + 4² = C²

3² is 3 x 3, which is 9

4² is 4 x 4, which is 16

  so

9 + 16 = C²

  so 

25 = C²

so
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 (3² + 4² = 5²).  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

 

Answer 3

Hi Joey;

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

Therefore, the equation we need is...

x-x₁

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