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Math Help - Distance Between Two Points

  1. #1
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    Distance Between Two Points

    Use the distance formula to find the distance between the two points
    (a, b) (b, a).
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    Quote Originally Posted by sologuitar View Post
    Use the distance formula to find the distance between the two points
    (a, b) (b, a).
    d = \sqrt{(b-a)^2 + (a-b)^2}

    all set up for you ... finish it.
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  3. #3
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    A(a,b) and B(b,a)
    The distance between the two points A and B:
    |\vec{AB}|=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}=\sqrt{(b-a)^2+(a-b)^2}=\sqrt{2}.|a-b|
    I don't know is it true or false ? So if it's false, please tell me !
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  4. #4
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    Quote Originally Posted by lovemath View Post
    A(a,b) and B(b,a)
    The distance between the two points A and B:
    |\vec{AB}|=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}=\sqrt{(b-a)^2+(a-b)^2}=\sqrt{2}.|a-b|
    I don't know is it true or false ? So if it's false, please tell me !
    That is true. Notice (a-b)^2=(b-a)^2 and \sqrt{(a-b)^2}=|a-b|=|b-a|
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    lovemath

    Username lovemath got the right answer. I tried getting the right answer over and over again but ended up with something totally different. How does the answer become sqrt{2}*|a - b|? In the radicand we have (a - b)^2 + (b - a)^2, which equals 2a^2 - 4ab +2b^2


    Last edited by sologuitar; February 2nd 2010 at 08:50 AM. Reason: Adding more text
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    Quote Originally Posted by sologuitar View Post
    Username lovemath got the right answer. I tried getting the right answer over and over again but ended up with something totally different. How does the answer become sqrt{2}*|a - b|? In the radicand we have (a - b)^2 + (b - a)^2, which equals 2a^2 - 4ab +2b^2

    Yes, and that is 2(a^2- 2ab+ b^2)= 2(a-b)^2. \sqrt{2(a-b)^2}= \sqrt{2}\sqrt{(a-b)^2} and \sqrt{(a-b)^2}= |a-b|
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    Yes but...

    Quote Originally Posted by HallsofIvy View Post
    Yes, and that is 2(a^2- 2ab+ b^2)= 2(a-b)^2. \sqrt{2(a-b)^2}= \sqrt{2}\sqrt{(a-b)^2} and \sqrt{(a-b)^2}= |a-b|
    I understand everything you said here. However, I don't understand how the absolute value comes into play.
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    Quote Originally Posted by sologuitar View Post
    I understand everything you said here. However, I don't understand how the absolute value comes into play.
    graph y = \sqrt{x^2} in your calculator ... what do you see?
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    Quote Originally Posted by sologuitar View Post
    I understand everything you said here. However, I don't understand how the absolute value comes into play.
    \sqrt{2^2}= \sqrt{4}= 2 \sqrt{(-2)^2}= \sqrt{4}= 2. Since \sqrt{x} is defined as "the non-negative number a such that a^2= x, if x is itself non-negative, then \sqrt{x^2}= x but if x is negative, \sqrt{x^2}= -x. In either case \sqrt{x^2}= |x|.
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    graphing

    Quote Originally Posted by skeeter View Post
    graph y = \sqrt{x^2} in your calculator ... what do you see?
    After graphing y = sqrt{x^2}, I got the line y = x, which crosses through the origin. What's your point?
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  11. #11
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    Quote Originally Posted by sologuitar View Post
    After graphing y = sqrt{x^2}, I got the line y = x, which crosses through the origin. What's your point?
    recheck your graph ...
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