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Math Help - crossing over line

  1. #1
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    Wink crossing over line

    Let l be a line through (x_0,y_0) cutting the line k: y=mx+c and forming an angle \alpha. Prove that the equation of the line l is
    y-y_0=\frac{m-\tan\alpha}{1+m\tan\alpha}(x-x_0) or
    y-y_0=\frac{\tan\alpha-m}{1+m\tan\alpha}(x-x_0)
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  2. #2
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    Hello GTK X Hunter
    Quote Originally Posted by GTK X Hunter View Post
    Let l be a line through (x_0,y_0) cutting the line k: y=mx+c and forming an angle \alpha. Prove that the equation of the line l is
    y-y_0=\frac{m-\tan\alpha}{1+m\tan\alpha}(x-x_0) or
    y-y_0=\frac{\tan\alpha-m}{1+m\tan\alpha}(x-x_0)
    I think you have an error in the second of these two equations. Here is what I believe to be the correct result.

    The equation of the line with gradient m_1 passing through the point (x_0,y_0) is
    y-y_0=m_1(x-x_0)
    This makes an angle \theta, say, with the direction of the x-axis, where \tan\theta = m_1.

    The line y = mx + c makes an angle \phi, say, with the x-axis, where \tan\phi = m.

    Then, depending on which side of the line y = mx + c the angle \alpha is made:
    \theta= \phi\pm\alpha (See the attached diagram. The lines shown are not necessarily the actual lines, but simply parallel to them through a single point.)

    \Rightarrow m_1=\tan\theta = \tan(\phi\pm\alpha)= \frac{\tan\phi \pm\tan\alpha}{1\mp\tan\phi\tan\alpha}

    =\frac{m\pm\tan\alpha}{1\mp m\tan\alpha}
    So the two gradients are \frac{m+\tan\alpha}{1-\tan\alpha} and \frac{m-\tan\alpha}{1+\tan\alpha}.

    The error in the question can be easily verified, because the gradient of one of the the given lines is (-1) \times the gradient of the other. Therefore they make equal angles above and below the x-axis. They should, of course, make equal angles above and below the line y = mx + c.

    Grandad
    Attached Thumbnails Attached Thumbnails crossing over line-untitled.jpg  
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  3. #3
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    Thankyou so much grandad
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