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Thread: Find an equation for the line

  1. #1
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    Post Find an equation for the line

    Hi! I am trying to study for my final and am really stuck on a problem. I'm not really sure how to start this one off--

    A line goes through the point (p,q) where p satisfies (1/p - 2/3) over (1/4-1/3) = 7 and q is the imaginary part of 2-3i over 4+5i. The slope of the line is the extraneous solution to x + the square root of (x+1)=5. Find an equation for the line (No caculator approximations. Do exact work)

    Thank you in advance for your help... this is kind of a tough one!

    Ashley
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  2. #2
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    I can help you find p & q


    firstly p:

    $\displaystyle \frac{\frac{1}{p}-\frac{2}{3}}{\frac{1}{4}-\frac{1}{3}}=7$

    $\displaystyle \frac{1}{p}-\frac{2}{3} = 7(\frac{1}{4}-\frac{1}{3})$

    $\displaystyle \frac{1}{p}-\frac{2}{3} = 7(\frac{3}{12}-\frac{4}{12})$

    $\displaystyle \frac{1}{p}-\frac{2}{3} = \frac{-7}{12}$

    $\displaystyle \frac{1}{p} = \frac{-7}{12}+\frac{2}{3}$

    $\displaystyle \frac{1}{p} = \frac{-7}{12}+\frac{8}{12}$

    $\displaystyle \frac{1}{p} = \frac{1}{12}$

    $\displaystyle p=12$

    now q:

    $\displaystyle q=Im(\frac{2-3i}{4+5i})$

    we need to find the imaginary part of $\displaystyle \frac{2-3i}{4+5i}$ multiply through by the complex conjugate.

    $\displaystyle \frac{2-3i}{4+5i}\times\frac{4-5i}{4-5i}$

    $\displaystyle \frac{8-10i-12i+15i^2}{16-25i^2}$

    $\displaystyle \frac{8-22i-15}{16+25}$

    $\displaystyle \frac{-7-22i}{41}$

    therefore $\displaystyle q= Im(\frac{-7-22i}{41}) = \frac{-22i}{41}$
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  3. #3
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    Woah.... I never would have gotten that far! Thank you so much for getting me started. This may sound like a dumb question, but when it says find an equation for the line- does that mean in y=mx+b formula. I just don't understand where p and q fit in or how you can have a slope that is an extraneous solution... this final is going to get me

    Thanks for the help!
    Ashley
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  4. #4
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    Without seeing the actual question or knowing the context it was asked in I would assume that the line does have the model $\displaystyle y=mx+b$

    And we now have the following information (assuming its correct!)

    $\displaystyle (p,q) = (x,y) = (12, \frac{-22i}{41})$ and $\displaystyle m= 7$

    Substituting these into your model we can solve for b and then have the equation.

    $\displaystyle y=mx+b$

    with $\displaystyle (p,q)=(x,y) =(12, \frac{-22i}{41})$ and $\displaystyle m= 7$

    gives $\displaystyle \frac{-22i}{41}=7\times 12+b$

    gives $\displaystyle \frac{-22i}{41}=84+b$

    gives $\displaystyle b = \frac{-22i}{41}-84$

    so the linear model is

    $\displaystyle y=7x+\left(\frac{-22i}{41}-84\right)$

    or in context of the problem

    $\displaystyle q=7p+\left(\frac{-22i}{41}-84\right)$

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  5. #5
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    Smile

    You have no idea how much that helps!!! I think I can make myself understand this now! You have done a great job of explaining this... maybe this final wont kill me after all haha

    Thank you!!!!!
    Ashley
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