Results 1 to 6 of 6

Thread: Coordinate geomtry

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    72

    Coordinate geomtry

    Hi, I have a problem that is giving me trouble solving. The problem reads The vertices of a $\displaystyle \triangle$ ABC are at the points A$\displaystyle (a, 0)$, B$\displaystyle (0, b)$, C$\displaystyle (c, d)$. If $\displaystyle \angle$B =$\displaystyle 90 degrees$. find a relaionship between a, b, c and d. The book answere is $\displaystyle b(d-b)=ac$. Can someone help me to fugure out this proble please thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Jan 2010
    Posts
    13
    It is simple. You gotta tally lengths:
    $\displaystyle |AB|=\sqrt{a^2 + b^2}$
    $\displaystyle |BC|=\sqrt{c^2+ (d-b)^2}$
    $\displaystyle |AC|= \sqrt{(c-1)^2 + d^2}$

    From Pitagoras' theorem:
    $\displaystyle |AB|^2 + |BC|^2 = |AC|^2$

    After simple transformation you should get:
    $\displaystyle b^2 - bd + ca=0$
    And finally the answer you wrote.
    Last edited by Oloria; Jan 12th 2010 at 11:17 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2009
    Posts
    72
    Quote Originally Posted by Oloria View Post
    It is simple. You gotta tally lengths:
    $\displaystyle |AB|=\sqrt{a^2 + b^2}$
    $\displaystyle |BC|=\sqrt{x^2+ (d-b)^2}$
    $\displaystyle |AC|= \sqrt{(c-1)^2 + d^2}$

    From Pitagoras' theorem:
    $\displaystyle |AB|^2 = |BC|^2 = |AC|^2$

    After simple transformation you should get:
    $\displaystyle b^2 - bd + ca=0$
    And finally the answer you wrote.
    Thank you very much. When i add the results I am having trouble comming up with the answer. Are you suppose to just add the results?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jan 2010
    Posts
    13
    Show me your solution, please. Then I will point you to error.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2009
    Posts
    72
    Quote Originally Posted by Oloria View Post
    Show me your solution, please. Then I will point you to error.
    Hi, this is what I got so far
    $\displaystyle a^2 + b^2 +c^2-(d-b)^2=(c-1)^2+d^2$

    $\displaystyle
    a^2-b^2-2b(-d)-(-d)^2+b^2+c^2+2c(-1)+(-1)^2+c^2+d^2$

    $\displaystyle
    a^2+2c^2+2bd-2c+1
    $
    I have a feeling that I am on the wrong track.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jan 2010
    Posts
    13
    Quote Originally Posted by scrible View Post
    Hi, this is what I got so far
    $\displaystyle a^2 + b^2 +c^2-(d-b)^2=(c-1)^2+d^2$
    On the left side you subtract (d-b)^2 and you should sum it. Correct:
    $\displaystyle a^2 + b^2 +c^2+(d-b)^2=(c-1)^2+d^2$

    I have a feeling that I am on the wrong track.
    Wrong feeling. I did it to the end and my result is the same as in the book.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. circle geomtry III
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Aug 21st 2010, 07:55 AM
  2. circle geomtry II
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Aug 21st 2010, 07:44 AM
  3. geomtry
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 5th 2009, 03:04 PM
  4. Geomtry
    Posted in the Geometry Forum
    Replies: 5
    Last Post: Mar 18th 2009, 02:25 PM
  5. Analytic Geomtry
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Oct 10th 2006, 12:53 PM

Search Tags


/mathhelpforum @mathhelpforum