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Math Help - Challenging coordinate geometry?

  1. #1
    Super Member fardeen_gen's Avatar
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    Challenging coordinate geometry?

    Find the correct statement(s)?

    A) the lines x + log b(base a).y + log c(base a) = 0, log a(base b).x + y + log c(base b)= 0 and log a(base c).x + log b(base c).y + 1 = 0 are concurrent

    B)Equation of a straight line passing through origin and making with x axis an angle twice the size of the angle made by the line 5y = x with the x-axis is 5y = 2x

    C) The triangle formed by 2x - y + 5 = 0, x + y - 5 = 0 and x - 2y - 5 = 0 is isoceles

    D) None of these.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by fardeen_gen View Post
    Find the correct statement(s)?

    A) the lines x + log b(base a).y + log c(base a) = 0, log a(base b).x + y + log c(base b)= 0 and log a(base c).x + log b(base c).y + 1 = 0 are concurrent

    B)Equation of a straight line passing through origin and making with x axis an angle twice the size of the angle made by the line 5y = x with the x-axis is 5y = 2x

    C) The triangle formed by 2x - y + 5 = 0, x + y - 5 = 0 and x - 2y - 5 = 0 is isoceles

    D) None of these.
    B) False. Check the arctan of their slopes. Almost, but not twice the size

    \tan^{-1}\left(\frac{1}{5}\right)\approx 11.3^{\circ}

    \tan^{-1}\left(\frac{2}{5}\right)\approx 21.8^{\circ}


    C) True. The triangle is isosceles

    2x-y+5=0 \ \ and \ \ x+y-5=0 intersect at A(0, 5)

    2x-y+5=0 \ \ and \ \ x-2y-5=0 intersect at B(-5, -5)

    x+y-5=0 \ \ and \ \ x-2y-5=0 intersect at C(5, 0)

    Using the distance formula, it can be shown that AB=BC=\sqrt{125}.
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  3. #3
    Super Member fardeen_gen's Avatar
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    Thank you! I figured out C just now. But what about A? I am not familiar with logarithmic quantities to a great extent.
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  4. #4
    A riddle wrapped in an enigma
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    Quote Originally Posted by fardeen_gen View Post
    Find the correct statement(s)?

    A) the lines x + log b(base a).y + log c(base a) = 0, log a(base b).x + y + log c(base b)= 0 and log a(base c).x + log b(base c).y + 1 = 0 are concurrent
    x+(\log_a b)y + \log_a c =0

    (\log_b a)x+y+\log_b c=0

    (\log_c a)x+(\log_c b)y+1=0

    I'm not sure about this one, either. I suggest you post it as a separate thread and see if any of our other astute tutors can tackle it. I hope I translated your code correctly.
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  5. #5
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    Hello, fardeen_gen!

    Ah, masters explained (B) and (C) . . . good!


    Find the correct statement(s).
    A very challenging problem!

    A) The lines: . \begin{Bmatrix}x + (\log_ab)y &=& -\log_ac & [1] \\ (\log_ba)x + y &=& -\log_bc & [2]\\<br />
(\log_ca)x + (\log_cb)y &=&  -1 & [3] \end{Bmatrix}\quad\text{ are concurrent.} . . . . True
    Using the Base-Change Formula:

    [1]\;\;x + \left(\frac{\log b}{\log a}\right)y \:=\:-\frac{\log c}{\log a} \quad\Rightarrow\quad (\log a)x + (\log b)y \:=\:-\log c

    [2]\;\;\left(\frac{\log a}{\log b} \right)x + y \:=\:-\frac{\log c}{\log b} \quad\Rightarrow\quad (\log a)x + (\log b)y \:=\:-\log c

    [3]\;\;\left(\frac{\log a}{\log c}\right)x + \left(\frac{\log b}{\log c}\right)y \:=\:-1 \quad\Rightarrow\quad (\log a)x + (\log b)y \:=\:-\log c


    The lines are identical . . . They are concurrent.
    . . (They share a common point.)

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