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Math Help - Triangle problem in coordinate geometry

  1. #16
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    Re: Triangle problem in coordinate geometry

    Hmm, something you could use: \tan\left(\dfrac{\pi}{2}-A\right) = \cot A. So, you could use your formula to get the tangent of angle \dfrac{\pi}{2}-A using m_2 and the slope of the hypotenuse.
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  2. #17
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    Re: Triangle problem in coordinate geometry

    no! i did this right. A is the angle between the hypotenuse whose slope is 2/5 and m1.
    then is used tan(90-A)= 1/tan A for m2 and and replaced m2 with -1/m2. to get two values of m1 with respect to tan A.
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  3. #18
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    Re: Triangle problem in coordinate geometry

    Then, here are two possible reasons for the result:

    1. You did not take into account the absolute value. Take the negative of one of those terms. Then you will probably get 0=0.

    2. You are dividing by zero somewhere. Note that in the formula for tangent, if m_1m_2 = -1, the denominator is zero. Perhaps that can happen elsewhere in the problem, as well.

    Edit: I just calculated it out. You did not take into account the absolute value. Calculating \tan A with m_1 and \dfrac{2}{5}) or calculating \dfrac{1}{\tan A} with m_2 = -\dfrac{1}{m_1} and \dfrac{2}{5} and solving for m_1 each time, I get the exact same result:

    m_1 = \dfrac{5\tan A \pm 2}{\pm 5 - 2\tan A} yielding the possible outcomes m_1 = \dfrac{5\tan A + 2}{5-2\tan A} or m_1 = \dfrac{2-5\tan A}{5+2\tan A}. These cannot BOTH be true unless \tan A = 0.
    Last edited by SlipEternal; November 17th 2013 at 09:31 AM.
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  4. #19
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    Re: Triangle problem in coordinate geometry

    The absolute value in this formula is used to find the tanA of the acute angle as the acute angle can be formed by the lines in two ways. But as this is a right triangle. I dont think we need to take the modulus into account.
    Last edited by AaPa; November 17th 2013 at 10:40 AM.
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  5. #20
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    Re: Triangle problem in coordinate geometry

    \tan A = \pm \dfrac{\tfrac{2}{5} - m_1}{1+\tfrac{2}{5}m_1} = \pm \dfrac{2-5m_1}{5+2m_1}

    5\tan A + 2m_1\tan A = \pm (2-5m_1) = \pm 2 \mp 5m_1

    2m_1\tan A \pm 5m_1 = \pm 2 - 5 \tan A

    (2\tan A \pm 5)m_1 = \pm 2-5\tan A

    m_1 = \dfrac{\pm 2-5\tan A}{2\tan A \pm 5}
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  6. #21
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    Re: Triangle problem in coordinate geometry

    The absolute value in this formula is used to find the tanA of the acute angle as the acute angle can be formed by the lines in two ways. But as this is a right triangle. I dont think we need to take the modulus into account.
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  7. #22
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    Re: Triangle problem in coordinate geometry

    If m_1 > \dfrac{2}{5} or if m_1<-\dfrac{5}{2}, then  \dfrac{\tfrac{2}{5}-m_1}{1+\tfrac{2}{5}m_1}<0. So, the plus/minus signs are needed. I don't understand what you mean by "take the modulus into account". What modulus?
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  8. #23
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    Re: Triangle problem in coordinate geometry

    m1 is making an acute angle A with the line of slope 2/5. Doesn't this mean tan A is positive?
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  9. #24
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    Re: Triangle problem in coordinate geometry

    Yes, so if -\dfrac{5}{2} < m_1 < \dfrac{2}{5}, you would use a plus sign for, otherwise, a minus sign. This way, \tan A will be positive.
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  10. #25
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    Re: Triangle problem in coordinate geometry

    Yes but when it makes an acute angle doesn't it automatically mean that m1 lies between -5/2 and 2/5 ?
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  11. #26
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    Re: Triangle problem in coordinate geometry

    When two lines intersect, they form four angles. Unless they are all right angles, two of them will be obtuse and two will be acute. That is why the formula you are given takes the absolute value. It cannot be ignored.
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  12. #27
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    Re: Triangle problem in coordinate geometry

    Here is a better way to explain it. Suppose m_1=-1. That is a possible value where the formula is positive. Then m_2=1 which is not within the range that would allow you to use positive values. In fact, given any -\dfrac{5}{2} < m_1 < \dfrac{2}{5}, m_2<-\dfrac{5}{2} or \dfrac{2}{5} < m_2
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  13. #28
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    Re: Triangle problem in coordinate geometry

    oh thank you! I got it. The two lines are making both acute and obtuse angles with each other. foolish me. thanks a lot for helping me out.
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