# A point reflected over a line?

• Sep 22nd 2010, 05:29 PM
hashshashin715
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?
• Sep 22nd 2010, 07:07 PM
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Quote:

Originally Posted by hashshashin715
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
• Sep 23rd 2010, 02:52 AM
hashshashin715
Ok, thanks, I'll keep vectors in mind when I do this.
• Sep 23rd 2010, 03:15 AM
HallsofIvy
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.