Results 1 to 15 of 15

Math Help - Coordinate geometry

  1. #1
    Member
    Joined
    Jul 2007
    Posts
    147

    Coordinate geometry

    Dear forum members,


    could you please help me.


    Find all possible values of a and b given that y=ax+14 is the perpendicular bisector of the line joining (1,2) to(b,6) .



    I have tried everything(except for the right method of course), but I always end up with something illogical.

    Thank you in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    First of all, find the equation of the line between A(1,2) and B(b,6) (by solving y=mx+n)

    Then, let H be the intersection of the two lines.

    H is the middle of [AB]. This may give you a relation between a and b.


    For the second one, you can take a random point C on y=ax+14 (in function of a) and say that CA=CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2007
    Posts
    147
    Thank you, Moo, for answering to my post again.





    I found the coordinates of the midpoint


    X=(1+b)/2 and Y= 4

    the equation for line which passes through points (1,2) and (b,6) is

    Y= \frac{4}{b-1}x-(\frac{4}{b-1}+\frac{2b-2}{b-1})





    then wouldn't a be the inverse negative number of the gradient of the first equation?



    I have tried plugging x and y in the equation of the first line and then solving for b, but I have never gotten a logical answer.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Perfect for the midpoint

    For the equation of the line, i find something else :

    n=2-m=2-4/(b-1)

    Y=\frac{4}{b-1} x + (\frac{2b-2}{b-1} - \frac{4}{b-1})

    Then, take the point at random of abscissa (b-1) (so that it will simplify for its y). This point is at equal distance of A and B.


    Can you continue the exercise ? It's apparently not sufficient, but i come back in 30 minutes


    You could also act with direction vectors, but i don't know if you've studied it...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jul 2007
    Posts
    147
    Quote Originally Posted by Moo View Post
    Perfect for the midpoint

    For the equation of the line, i find something else :

    n=2-m=2-4/(b-1)

    Y=\frac{4}{b-1} x + (\frac{2b-2}{b-1} - \frac{4}{b-1})

    Then, take the point at random of abscissa (b-1) (so that it will simplify for its y). This point is at equal distance of A and B.


    Can you continue the exercise ? It's apparently not sufficient, but i come back in 30 minutes


    You could also act with direction vectors, but i don't know if you've studied it...

    Thank you Moo for replying. But can you tell me why I just can't plug the coordinates of the midpoint into that equation, so 4=\frac{4}{b-1} *\frac{1+b}{2}+ (\frac{2b-2}{b-1} - \frac{4}{b-1})





    I mean the coordinates of the midpoint have to satisfy the equation of a line, or am I totally wrong?

    And I have never studied direction vectors, so I have no idea what they are.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Well, with my formula (which is a bit different from yours), i find it exact. When you factorised the - sign, you made a little mistake. Your idea of verifying is very good
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jul 2007
    Posts
    147
    Quote Originally Posted by Moo View Post
    Well, with my formula (which is a bit different from yours), i find it exact. When you factorised the - sign, you made a little mistake. Your idea of verifying is very good


    Thank you, and feel very stupid for saying this, I still don't get how to do it?


    what did you get for your answer. According to my book the answers should be a=-2, b=9 or a= 5/2 and b=-9



    Thank you so much for giving up your time to help me.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    I'm here to help :-)

    I don't have the results, but i know the steps to go to it ^^

    You wrote Y= \frac{4}{b-1}x-(\frac{4}{b-1}+\frac{2b-2}{b-1}) , which is [Math]Y=\frac{4}{b-1}x-\frac{4}{b-1}-\frac{2b-2}{b-1}[/tex]

    But it's +\frac{2b-2}{b-1}, not -

    And you can simplify : (2b-2)-4=2b-6
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Ok, let's take it again from the beginning :

    - you got the coordinates of H((b+1)/2 ; 4), which is on the two lines, including y=ax+14. If you replace x and y in this equation, you will have a with an expression including b -> tell me what you have

    - i'm really sorry, but i think finding the last equation was unnecessary (at least, you've been training )

    - choose a random point on the line y=ax+14, for example C(0;14) (simple, because it deletes a). Then, write the equation : CA=CB (C is on the perpendicular bisector of [AB])

    AC=\sqrt{(x_C-x_A)^2+(y_C-y_A)^2}

    BC=\sqrt{(x_B-x_C)^2+(y_B-y_C)^2}

    This will give you two possible values for b, and you have a with a previous relation.

    I've checked, you'll get the values given by the correction
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Jul 2007
    Posts
    147
    this is what I get when trying to solve for b, and it all cancels out


    8b^2-16b+8=8b^2-16b+8
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Read my last message :-)

    I'm really sorry to have put you on a wrong way
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Jul 2007
    Posts
    147
    but if I substitute the coordinates into Y=ax+14, I will get -20=a+b
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Not really ;-)

    H((b+1)/2 ; 4)

    So in y=ax+14 :

    4=a(b+1)/2 + 14

    => -20=a(b+1)

    => a=-20/(b+1) :-)



    H is on this line, because it's the midpoint of [AB] and y=ax+14 is the perpendicular bisector.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member
    Joined
    Jul 2007
    Posts
    147
    Quote Originally Posted by Moo View Post
    Not really ;-)

    H((b+1)/2 ; 4)

    So in y=ax+14 :

    4=a(b+1)/2 + 14

    => -20=a(b+1)

    => a=-20/(b+1) :-)



    H is on this line, because it's the midpoint of [AB] and y=ax+14 is the perpendicular bisector.


















    WOW!


    You're math skills are amazing, I got the right answer now!


    Thank you so much for helping me, I cannot say how much it means to me that someone has devoted so much of their time to voluntarily help someone else.




    Though I think I am going to drop math next year, I just can't do stuff like this, and if I'm correct this is not so advanced for high school.









    Have a good night, and thank you for everything!
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    You're welcome, it was a pleasure ^^
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Coordinate Geometry
    Posted in the Geometry Forum
    Replies: 1
    Last Post: February 15th 2011, 11:06 PM
  2. Coordinate Geometry
    Posted in the Geometry Forum
    Replies: 1
    Last Post: June 5th 2010, 04:21 AM
  3. coordinate geometry
    Posted in the Algebra Forum
    Replies: 6
    Last Post: November 25th 2009, 07:28 AM
  4. coordinate geometry
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: November 24th 2009, 09:35 AM
  5. Coordinate Geometry
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 12th 2008, 04:26 PM

Search Tags


/mathhelpforum @mathhelpforum