Results 1 to 7 of 7

Math Help - Distance of Coordinates

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    5

    Distance of Coordinates

    I'm really weak at maths and I have to study for my upcoming exams. I can't seem to find the distance of these coordinates.

    1. (p,q) and (q,p)
    2. (p+4q, p-q) and (p-3q, p)

    I think in the first one I have to open the brackets using the a2-2ab+b2 formula and in the second question, I am doing it normally but my answer stops at q2+q2 under square root. Help?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2009
    Posts
    263

    Smile

    Quote Originally Posted by want2math View Post
    I'm really weak at maths and I have to study for my upcoming exams. I can't seem to find the distance of these coordinates.

    1. (p,q) and (q,p)
    2. (p+4q, p-q) and (p-3q, p)

    I think in the first one I have to open the brackets using the a2-2ab+b2 formula and in the second question, I am doing it normally but my answer stops at q2+q2 under square root. Help?
    the distance for points (x_{1},y_{1}) and (x_{2},y_{2}) : d = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}

    1. d = \sqrt{(q - p)^2 + (p - q)^2} =  \sqrt{(q - p)^2 + (q - p)^2} =  \sqrt{2(q - p)^2} = (q - p) \sqrt{2}

    try number 2
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2010
    Posts
    5
    Quote Originally Posted by dedust View Post
    the distance for points (x_{1},y_{1}) and (x_{2},y_{2}) : d = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}

    1. d = \sqrt{(q - p)^2 + (p - q)^2} =  \sqrt{(q - p)^2 + (q - p)^2} =  \sqrt{2(q - p)^2} = (q - p) \sqrt{2}

    try number 2
    \sqrt{(p-3q-p+4q)^2 + (p-p-q)^2}
    \sqrt{(p-p+4q-3q)^2 + (p-p-q)^2}
    \sqrt{q^2 + q^2}

    What am I doing wrong?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2009
    Posts
    263

    Smile

    Quote Originally Posted by want2math View Post
    \sqrt{(p-3q-p+4q)^2 + (p-p-q)^2}
    \sqrt{(p-p+4q-3q)^2 + (p-p-q)^2} \leftarrow should be \sqrt{(p-p-4q-3q)^2 + (p-p+q)^2}


    What am I doing wrong?
    . . .
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,370
    Thanks
    1314
    Quote Originally Posted by want2math View Post
    \sqrt{(p-3q-p+4q)^2 + (p-p-q)^2}
    \sqrt{(p-p+4q-3q)^2 + (p-p-q)^2}
    \sqrt{q^2 + q^2}

    What am I doing wrong?
    (p+4q)- (p- 3q)= (p-p)+ (4q+ 3q)= 7q
    (p- q)- (p)= -q but (-q)^2= q^2.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by want2math View Post
    \sqrt{(p-3q-p+4q)^2 + (p-p-q)^2}
    \sqrt{(p-p+4q-3q)^2 + (p-p-q)^2}
    \sqrt{q^2 + q^2}

    What am I doing wrong?
    It's easier to follow if you use brackets to show which term is which. I can't follow that above so I'll detail my way below

    It seems you're getting confused with negative signs. a-(a-b) = a-a+b = b. Remember subtracting a negative is like adding a positive

    \sqrt{[(p-3q)-(p+4q)]^2 + [p-(p-q)]^2}

    (p-3q)-(p+4q) = p-3p-p-4q = -7q

    p-(p-q) = p-p+q = q

    \therefore \: \: \sqrt{[(p-3q)-(p+4q)]^2 + [p-(p-q)]^2} = \sqrt{(-7q)^2 + q^2} = \sqrt{50q^2} = \sqrt{2 \cdot 25 \cdot q^2}

    25 and q^2 are both squares so we can remove them from the surd: 5q \sqrt{2}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jan 2010
    Posts
    5
    Quote Originally Posted by e^(i*pi) View Post
    It's easier to follow if you use brackets to show which term is which. I can't follow that above so I'll detail my way below

    It seems you're getting confused with negative signs. a-(a-b) = a-a+b = b. Remember subtracting a negative is like adding a positive

    \sqrt{[(p-3q)-(p+4q)]^2 + [p-(p-q)]^2}

    (p-3q)-(p+4q) = p-3p-p-4q = -7q

    p-(p-q) = p-p+q = q

    \therefore \: \: \sqrt{[(p-3q)-(p+4q)]^2 + [p-(p-q)]^2} = \sqrt{(-7q)^2 + q^2} = \sqrt{50q^2} = \sqrt{2 \cdot 25 \cdot q^2}

    25 and q^2 are both squares so we can remove them from the surd: 5q \sqrt{2}
    Thanks for the amazing post.

    This post clears any doubt in my mind.

    Again, thank you all.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: September 24th 2010, 04:33 AM
  2. Distance between Coordinates
    Posted in the Geometry Forum
    Replies: 2
    Last Post: December 14th 2009, 11:20 PM
  3. Replies: 2
    Last Post: July 28th 2009, 01:00 AM
  4. Distance b. Points (Polar Coordinates)
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: February 19th 2009, 12:49 AM
  5. Replies: 6
    Last Post: February 4th 2009, 12:12 AM

Search Tags


/mathhelpforum @mathhelpforum