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Math Help - line of symmetry

  1. #1
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    line of symmetry

    having trouble with question 8 on this paper:

    http://www.mei.org.uk/files/papers/2012_Jan_c3.pdf


    im guessing its because the gradient at P is -1/3 and not -1 but i cant justify this. im thinking the gradient needs to be perpendicular to y=x bit can see no reason this helps
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  2. #2
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    Re: line of symmetry

    along the line y = x we have to have dy/dx = -1 if the curve is symmetric, do we not?

    think of it this way: suppose our curve WERE symmetric and (x_1,y_1) is a point of the curve below y = x, and (x_2,y_2) is the corresponding point of the curve above y = x.

    if the curve is symmetric, we have (x_2,y_2) = (y_1,x_1).

    now, what is the slope between those two points? it's:

    \frac{y_2-y_1}{x_2-x_1} = \frac{x_1 - y_1}{y_1 - x_1} = -1

    now suppose a point where the curve crosses y = x is (a,a). then as x_1 \to a, the slope between the two points approaches dy/dx, but this slope is constantly -1.
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  3. #3
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    Re: line of symmetry

    Quote Originally Posted by Deveno View Post
    along the line y = x we have to have dy/dx = -1 if the curve is symmetric, do we not?

    think of it this way: suppose our curve WERE symmetric and (x_1,y_1) is a point of the curve below y = x, and (x_2,y_2) is the corresponding point of the curve above y = x.

    if the curve is symmetric, we have (x_2,y_2) = (y_1,x_1).

    now, what is the slope between those two points? it's:

    \frac{y_2-y_1}{x_2-x_1} = \frac{x_1 - y_1}{y_1 - x_1} = -1

    now suppose a point where the curve crosses y = x is (a,a). then as x_1 \to a, the slope between the two points approaches dy/dx, but this slope is constantly -1.
    hi,thanks for the reply. so its nothing to do with perpendicular? what would the grad need to be if the it was symmetric about y=kx for k not 1?
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  4. #4
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    Re: line of symmetry

    well, yes, it does have to be perpendicular, but proving this for the line y = kx is a lot more troublesome (it's messy).

    if a curve is symmetric about the line y = kx, then the point corresponding to (x_1,y_1) below the line, is the point (above the line):

    (x_2,y_2) = \left(\frac{1-k^2}{1+k^2}x_1 + \frac{2k}{1+k^2}y_1,\ \frac{2k}{1+k^2}x_1 - \frac{1-k^2}{1+k^2}y_1\right)

    a similar (but more involved) calculation as before shows that:

    \frac{y_2-y_1}{x_2-x_1} = \frac{2kx_1-2y_1}{2ky_1-2k^2x_1} = \frac{1}{k}\cdot\frac{kx_1 - y_1}{y_1-kx_1} = -\frac{1}{k}

    which is perpendicular to the line.

    intuitively, however, you can see this by rotating the line by arctan(k) - π/4
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