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Math Help - A point reflected over a line?

  1. #1
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    A point reflected over a line?

    Lets say we have a line that goes through the origin, which we'll write as y = ax. If I have a point (x, y) and reflect it over the line y = ax, then what is the resulting point?

    For example, when we reflect over x = 0, the outcome is (-x, y)
    Or when we reflect over y = x, the outcome is (y, x)

    So is there a general way of writing the reflected point over the line y = ax?
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  2. #2
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    Quote Originally Posted by hashshashin715 View Post
    Lets say we have a line that goes through the origin, which we'll write as y = ax. If I have a point (x, y) and reflect it over the line y = ax, then what is the resulting point?

    For example, when we reflect over x = 0, the outcome is (-x, y)
    Or when we reflect over y = x, the outcome is (y, x)

    So is there a general way of writing the reflected point over the line y = ax?
    Note that it is impossible to express the line x=0 this way.

    More general way to express a line through the origin is ax + by = 0.

    Alternatively we can use an angle, say 0\le\theta < \pi.

    Probably easiest is to think in terms of vectors. Let O be the origin, let L be a line through O, let P be a point, and let P' be the reflection of P across L. Note that P and P' have the same distance from the origin (same magnitude), and note that connecting O, P, and P' either gives you a line segment or an isosceles triangle..

    Edit: You might find this useful, particularly the intro

    http://planetmath.org/encyclopedia/D...ionMatrix.html
    Last edited by undefined; September 22nd 2010 at 07:19 PM.
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  3. #3
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    Ok, thanks, I'll keep vectors in mind when I do this.
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  4. #4
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    To find the reflection of the point (x_0, y_0) in the line y= ax, geometrically, you draw a line, through (x_0, y_0), perpendicular to y= ax, then mark the desired point on that perpendicular line the same distance from the line as (x_0, y_0) but on the other side.

    Algebraically, the line through (x_0, y_0), perpendicular to y= ax, is given by y= -\frac{1}{a}(x- x_0)+ y_0. It will intersect y= ax where [tex]y= ax= -\frac{1}{a}(x- x_0)+ y_0[tex] or (a+ \frac{1}{a})x= \frac{a^2+ 1}{a}x= \frac{x_0}{a}+ y_0. x= \frac{x_0}{a^2+ 1}+ \frac{ay_0}{a^2+ 1}= \frac{x_0+ ay_0}{a^2+ 1} so y= \frac{ax_0+ a^2y_0}{a^2+ 1} (since it is on the line y= ax). x_0- x= \frac{a}{a^2+ 1}(ax_0+ y_0) and [tex]y_0- y= -\frac{1}{a^2+ 1}(ax_0+ y_0) so the square of the distance from the original point (x_0, y_0) to the line is \frac{a^2}{(a^2+ 1)^2}(ax_0+y_0)^2+ \frac{1}{(a^2+ 1)^2}(ax_0+ y_0)^2 = \frac{a^2+ 1}{(a^2+ 1)^2}(ax_0+ y_0)^2= \frac{(ax_0+ y_0)^2}{a^2+ 1}.

    The reflection of (x_0, y_0) in the line y= ax is the point of intersection of the circle (x- \frac{x_0+ ay_0}{a^2+ 1})^2+ (y- \frac{ax_0+a^2y_0}{a^2+ 1})^2= \frac{ax_0+ y_0)^2}{a^2+ 1} and the line y= -\frac{1}{a}(x- x_0)+ y_0. That will result in a quadratic equation with two solutions. One is the point (x_0, y_0) itself and the other is the reflected point.
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