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Math Help - [SOLVED] |AP BP| is minimum for...?

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] |AP BP| is minimum for...?

    Let A(-2, -1) & B(3, 2) be two fixed points and P be a point on the line 3x + y = 1, for which location of P, |AP BP| is minimum?

    A) (2,-5)
    B) (0,1)
    C) (1, -2)
    D) none
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  2. #2
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    fardeen,

    P = (x,3x+1)
    AP = d(A,P) = \sqrt{(x + 2)^2 + (3x + 2)^2}
    BP = d(B,P) = \sqrt{(x - 3)^2 + (3x - 1)^2}

    Now that you have all of them in the same variable, just calculate |AP - BP| and find x minimum using derivatives. For x_{min} found, calculate y_{min} using the line equation.

    If you're not supposed to use calculus, then inform us, for then a different approach will be needed.
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  3. #3
    Super Member fardeen_gen's Avatar
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    No calculus. Only coordinate geometry and normal geometry.
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  4. #4
    Super Member fardeen_gen's Avatar
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    Also I couldn't understand how do we get P(x , 3x + 1). Can you please explain?
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  5. #5
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    fardeen,

    1) A point P is denoted by an ordered pair (x,y). Since your line equation is 3x + y = 1, then y = 1 - 3x (I obviously got the signs wrong in my last post, I'm sorry). Since you now know y in terms of x, then P = (x,1-3x).

    2) Now that I though a little further into the question, consider this approach: first notice that (AP - AB) is able to assume every value in the real line. So the minimum value of the modulus must be zero. This obviously happens when P is equidistant of both A and B.

    So your condition is:
    d(P,A) = d(P,B)
    \sqrt{(x + 2)^2 + (3x + 2)^2} = \sqrt{(x - 3)^2 + (3x - 1)^2}
    Now you square both sides:
    (x + 2)^2 + (3x + 2)^2 = (x - 3)^2 + (3x - 1)^2
    This is a quadratic equation. Solve it for x and you will find the value of x for which distances are equal, and because of such the difference is zero.

    Hope this helps,
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