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Thread: Analytic

  1. #1
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    Analytic

    Corners of a triangle ABC is A(-2,5),B(4,2) and C(x,0).For the
    perimeter of the triangle to be minimum what must be the value of x?

    MY WORK:
    I found the answer as -16/7 according to the book -24/7.

    I found the slope of AB as 3/2. For perimeter to be minimum A-B-C must
    be collinear. So -2/x+4=3/2,and x=-16/7

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  2. #2
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    Re: Analytic

    Quote Originally Posted by kastamonu View Post
    Corners of a triangle ABC is A(-2,5),B(4,2) and C(x,0).For the
    perimeter of the triangle to be minimum what must be the value of x?

    MY WORK:
    I found the answer as -16/7 according to the book -24/7.

    I found the slope of AB as 3/2. For perimeter to be minimum A-B-C must
    be collinear. So -2/x+4=3/2,and x=-16/7

    Looks like you missed a negative sign. Probably B should be (-4,2)? Otherwise, the slope of AB is $\dfrac{2-5}{4-(-2)} = \dfrac{-3}{6} = -\dfrac{1}{2}$.

    Then, $\dfrac{0-2}{x-4} = -\dfrac{1}{2}$ implies $-4=4-x$, so $x=8$.
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  4. #4
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    Re: Analytic

    Yes it is B(-4,2). I made a typo.
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    Re: Analytic

    If A(-2,5) and B(-4,2) then

    A-D-C must be collinear where D(-4,-2)

    so x=-24/7
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    Re: Analytic

    Why do we have to find symmetric of B to y-axis?
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  7. #7
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    Re: Analytic

    I meant x-axis.
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  8. #8
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    Re: Analytic

    We don't have to do it this way but this is an alternative way to solve the problem.

    since CB=CD, minimizing CA+CB is equivalent to minimizing CA+CD which occurs when ACD is a straight line

    slope of AD=7/2

    so slope of CA=-5/(x+2)=7/2

    solve for x
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  9. #9
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    Re: Analytic

    Many Thanks.
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