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Math Help - >>Coordinate Geometry: Find The Distance<<

  1. #1
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    >>Coordinate Geometry: Find The Distance<<

    yes im new so pls excuse if i do something wrong but,

    I have an exam in eight hours and im panicking because im really tired.
    Essentially, ive forgotten how to calculate distance in C/G (Coordinate Geometry) Puzzled with the first revision question already:

    Find the distance between R(-1, 5) and S(2, -1)

    the formula is: d = \sqrt{(X_{B} - X_{A})^2 + (Y_{B} - Y_{A})^2}

    Could someone please walk me through the process of working this out? thankyoou so much!

    The booklet says: RS = \sqrt{(2-(-1))^2 + (-1-5)^2}
    Last edited by rooobbbb; June 14th 2009 at 07:43 AM.
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  2. #2
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    Quote Originally Posted by rooobbbb View Post
    yes im new so pls excuse if i do something wrong but,

    I have an exam in eight hours and im panicking because im really tired.
    Essentially, ive forgotten how to calculate distance in C/G (Coordinate Geometry) Puzzled with the first revision question already:

    Find the distance between R(-1, 5) and S(2, -1)

    the formula is: d = \sqrt{(X_{B} - X_{A})^2 + (Y_{B} - Y_{A})^2}

    Could someone please walk me through the process of working this out? thankyoou so much!

     <br />
RS=\sqrt{(-1-2)^2+(-1-5)^2}<br />
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  3. #3
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    The distance formula - geometry's great illusion

    The distance formula is one of those equations which looks way more complicated than it is.

    1) Forget coordinates for a second. What we are really looking at is the Pythagorean theorem:
    a^2 = b^2 + c^2
    Or, rewritten:
    a = \sqrt{b^2 + c^2}

    We use this theorem to find the length of the hypotenuse in a right triangle. In order to use it, we need a length x and y between the two points you have.

    2) So, lets get the length x and y. So the question here is, how far apart are the points (-1,5) and (2, -1) in the x and y direction? For x, we look at the x coordinates -1 and 2.
    If these points are on a straight line, -1 and 2 are 3 units apart from one another or (2 - (-1)) = 3.
    If the y coordinates are on a straight line, 5 and -1 are 6 units apart from one another or (5 - (-1)) = 6.

    3) The distances 3 and 6 are our "a" and "b" values in the Pythagorean theorem. If you put all this mumbo-jumbo together you will get the "distance formula" that you found:
     s = \sqrt{a^2 + b^2}<br />
 = \sqrt{(2 - (-1))^2 + (5-(-1))^2}

    4) Simplifying, we get:
    s = \sqrt{3^2 + 6^2}<br />
 = \sqrt{9 + 36}<br />
 = \sqrt{45}
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