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  • Jan 18, 2011, 09:34 PM
    daddysredneck24
    Pre Calculus Homework
    I'm having trouble understanding how to get an answer for this problem, I got the answer (I looked in the back of the book for help, hoping It would help me) but I don't know how to get that answer... OH by the way its distance formula... etc..

    Problem:
    Find all poings having an x-coordinate of 2 whose distance from the point (-2, -1) is 5.

    the right answer is:
    92, 2); (2, -4)
  • Jan 18, 2011, 10:40 PM
    Unknown008

    Let a point have the coordinates (x, y) which you don't know.

    However, you know that the x coordinate is 2, hence, the coordinates of the point(s) become (2, y).

    Now, the distance formula is:



    This distance is 5, x = 2, y =? xo = -2, yo = -1. Substitute those into the distance formula:



    Can you complete it now? Solve for y!
  • Jan 19, 2011, 02:52 PM
    daddysredneck24
    Thanks so much... but I'm still a little bit confused, every time I try to solve it I end up with a weird answer.. .
  • Jan 19, 2011, 11:12 PM
    Unknown008

    Would you be so kind as to show what you did? We can help you stop your mistakes so that you know where to be more careful next time :)
  • Jan 19, 2011, 11:54 PM
    daddysredneck24
    I finally figured out half of it I think. Here's what I put:

    5 =sqrt(2-(-2))^2+(y-(-1))^2)
    5 =sqrt(4)^2+(y+1)^2
    5 =sqrt(16+y^2+2y+1)
    5 =sqrt(17+y^2+2y)
    25 =17+y^2+2y -------- (I squared both sides to cancel out the sqrt)
    8 =y^2+2y
    4y =y^2
    y=2
    (2,2)
    that's how I figured it... Please correct my mistakes if you can
  • Jan 20, 2011, 12:01 AM
    Unknown008

    Those are good procedures up to this line:



    This is a quadratic equation, and so, it cannot become



    This, instead becomes:



    Then, factorise:



    This done, you get:

    or

    This gives you two answers, which are what the given answers say (You already know the x coordinate, which is x = 2) :)
  • Jan 20, 2011, 06:47 PM
    daddysredneck24
    Comment on Unknown008's post
    OH! Ok! Thank you SO much... I completely understand now... thanks again!
  • Jan 20, 2011, 09:31 PM
    Unknown008

    You're welcome :)

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