Results 1 to 2 of 2

Math Help - RE: Could you proof it?

  1. #1
    Newbie
    Joined
    Aug 2009
    From
    Now studying at INTEC, UiTM
    Posts
    5

    RE: Could you proof it?

    Hai, everyone..once again, its me, Ah Chan. Sorry for I'm quite inactive this last few weeks because I got lot of problems banging my doors and so busy, rushing the assignment...anyways, thanks to everyone for my previous last questions. Erm, I got a proofing question here that had took me quite a long times yet I haven't fully understand the solution for it, mind to help me out...thanks a lot anyways

    Prove for distance from a point to line formula;

    First, show that the distance from O (0,0) to Ax+By+C=0 is lCl / (square root)[A^2 + B^2] by finding the area of the shaded triangle using two different altitudes and then equating the areas....(Hints: Draw a right-angle triangle with two points, [0,(-C/B)] and [(-C/A),0]


    Then, now translate (h,k) and the line Ax+By+C=0 by the same amount so that (h,k) moves to (0,0)...

    The answer should be d = lAh+Bk+Cl / (square root)[A^2+B^2]..but I just couldn't figured it out, please do help me..thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by Ah Chan View Post
    Hai, everyone..once again, its me, Ah Chan. Sorry for I'm quite inactive this last few weeks because I got lot of problems banging my doors and so busy, rushing the assignment...anyways, thanks to everyone for my previous last questions. Erm, I got a proofing question here that had took me quite a long times yet I haven't fully understand the solution for it, mind to help me out...thanks a lot anyways

    Prove for distance from a point to line formula;

    First, show that the distance from O (0,0) to Ax+By+C=0 is lCl / (square root)[A^2 + B^2] by finding the area of the shaded triangle using two different altitudes and then equating the areas....(Hints: Draw a right-angle triangle with two points, [0,(-C/B)] and [(-C/A),0]


    Then, now translate (h,k) and the line Ax+By+C=0 by the same amount so that (h,k) moves to (0,0)...

    The answer should be d = lAh+Bk+Cl / (square root)[A^2+B^2]..but I just couldn't figured it out, please do help me..thanks
    First I am not sure what area has to do with distance in the problem....

    Key point 1. When we talk about the distance between a point and a line it means the perpendicular distance.

    With that in mind here we go.


    We start with the line

    Ax+By+C=0 \iff y=-\frac{A}{B}x-\frac{C}{B}

    So the slope of this line is m=-\frac{A}{B}

    So the slope of the perpendicular line is m_\perp=\frac{B}{A}

    So the equation of the line is y=\frac{B}{A}x

    Now we want to find the intersection of these two lines so we set them equal.

    \frac{B}{A}x=-\frac{A}{B}x-\frac{C}{B} clearing the fractions we get

    B^2x=-A^2x-AC \iff x=\frac{-AC}{A^2+B^2}

    Now we can sub this into either equation of the line to get the y coordinate

    y=\frac{B}{A}\cdot \frac{-AC}{A^2+B^2}=\frac{-BC}{A^2+B^2}

    Now we can use the pythagorean theorem to find the distance between the two lines.

     \left( \frac{-AC}{A^2+B^2} \right)^2+\left( \frac{-BC}{A^2+B^2} \right)^2=d^2

    Simplifing we get

    \frac{C^2(A^2+B^2)}{(A^2+B^2)^2}=d^2

    \frac{|C|}{\sqrt{A^2+B^2}}=d

    This should get you started. Good luck.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: October 19th 2010, 10:50 AM
  2. Replies: 0
    Last Post: June 29th 2010, 08:48 AM
  3. [SOLVED] direct proof and proof by contradiction
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: February 27th 2010, 10:07 PM
  4. Proof with algebra, and proof by induction (problems)
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: June 8th 2008, 01:20 PM
  5. proof that the proof that .999_ = 1 is not a proof (version)
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: April 14th 2008, 04:07 PM

Search Tags


/mathhelpforum @mathhelpforum