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Math Help - Trigonometry prove question

  1. #1
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    Trigonometry prove question

    I should add this is desperate. The bracket after m and c just shows that the value of m and c are assumed to be different for each straight line.* means multiply.

    Let a and b be the straight lines with equations y=m(1)x +c and y=m(2)x +c where m(1)*m(2) is not equal to 0. Use appropriate trigonometric formula to prove that a and b are perpendicular if and only if m(1)m(2) =1.


    Please note that by appropriate- keep means to keep it as simple as possible- note this is high school maths so please try not to get too advance.
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  2. #2
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    Quote Originally Posted by kingkaisai2
    I should add this is desperate. The bracket after m and c just shows that the value of m and c are assumed to be different for each straight line.* means multiply.

    Let a and b be the straight lines with equations y=m(1)x +c and y=m(2)x +c where m(1)*m(2) is not equal to 0. Use appropriate trigonometric formula to prove that a and b are perpendicular if and only if m(1)m(2) =1.


    Please note that by appropriate- keep means to keep it as simple as possible- note this is high school maths so please try not to get too advance.
    Actually you mean that,
    m_1\cdot m_2=-1
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  3. #3
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    Let \theta represent the angle a line forms with the x-axis. Now, this is important, there are two possible angles the one on the right and left. We let \theta be the one on the right.

    There is a theorem, given a line y=mx+b then \tan \theta =m. This theorem is crucial in this proof.

    Let us have two lines. One y=m_1x+b_1 crosses the x-axis at M_1 and the other y=m_2x+b_2 crosses the x-axis at, M_2. Let P be the point of intersection of these two lines. Let us assume that they are perpendicular, meaning that \not < M_1PM_2=90^o. Let \theta_1 we the angle formed by the first line, and \theta_2 be the angle formed by the second line. Then, because we have a right triangle, \theta_2=\theta_1+90^o. Therefore, \tan \theta_1 \tan (90^o+\theta_1) can expressed as, -\tan \theta_1 \cot \theta _1 cuz, \tan (90^o+x)=-\cot x but because tangent and cotanget are reciprocals functions there product gives one thus,
    -\tan \theta_1 \cot \theta_1 =-1
    This means that,
    \tan \theta_1 \tan \theta_2=-1
    But, m_1=\tan \theta_1, m_2=\tan \theta_2
    Thus,
    m_1m_2=-1
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  4. #4
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    For the converse,
    --------
    Let us have two lines. One y=m_1x+b_1 crosses the x-axis at M_1 and the other y=m_2x+b_2 crosses the x-axis at, M_2. Let P be the point of intersection of these two lines. Let us assume that m_1m_2=-1. Let \theta_1 we the angle formed by the first line, and \theta_2 be the angle formed by the second line. Then, because we have a triangle, \theta_2=\theta_1+P. Where P=\not < M_1PM_2. Therefore, m_1m_2=-1 but, m_1=\tan \theta_1 and m_2=\tan \theta_2 thus,
    \tan \theta_1\tan (\theta_1+P)=-1
    Divide both sides by \tan \theta_1,
    \tan(\theta_1+P)=-\cot \theta_1 because you get a reciprocal. Now using the same reason as above you can write, \tan(\theta_1+P)=\tan(\theta_1+90^o)
    Thus,
    \theta_1+P=\theta_1+90^o
    Thus,
    P=90^o.
    Proof is complete
    -----------


    Note (not for you): I know I could have not divided both sides sides by tangent without assuming the line was non-vertical and I cannot conclude,
    x=y from \tan x=\tan y without denostrating this hold for angles 0\leq x\leq \pi, i.e. domain of triangle.
    So please do not bother me with these posts, I omitted that because that was high schools math.
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