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Math Help - Reflection in line

  1. #1
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    Reflection in line

    Find the cooridiantes of the reflection of the point (a , b) in the line  y = mx + c.

    I keep getting VERY messy answers to this question, anyone want to give me a few pointers?
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  2. #2
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    i would say that you need to supply what the reflection of the point is with respect to, ie the x axis, y axis, or the origin, or if it is a reflection of a,b with respect to that line you mention, not in the line. In the line sounds like it is a point within that line.
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  3. #3
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    Quote Originally Posted by pellikkan View Post
    i would say that you need to supply what the reflection of the point is with respect to, ie the x axis, y axis, or the origin, or if it is a reflection of a,b with respect to that line you mention, not in the line. In the line sounds like it is a point within that line.
    sorry the reflection is with respect to the line y=mx +c
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  4. #4
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    ok, you need to find the line that is perpendicular to that line you mention, but which also contains the point a,b, ....see math book.

    then you would need to find the point of the intersection of those two lines,....solve the linear system.

    then you would need to make that intersection point the average, or midpoint, of the two reflected points, but solve that equation for the unknown reflected point, say a', b'.

    i hope that helps!
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  5. #5
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    Quote Originally Posted by bobak View Post
    Find the cooridiantes of the reflection of the point (a , b) in the line  y = mx + c.

    I keep getting VERY messy answers to this question, anyone want to give me a few pointers?
    I assume this approach will work. It eliminates a lot of mess.

    First shift down (up) the line c units and the point down (up) c units. Thus, now you are reflecting (a,b-c)=(a,d) through y=mx. Suppose the line makes an angle of \theta, then rotate the line and the point by \theta, the point ends up in the new location (a',d') in such a way that the line ends up at y=x (meaning we rotate y=mx enough degrees so it becomes the line y=x). Now reflect (a',d') through y=x, which is simply (d',a'). Now you need to undo your steps. Meaning rotate (in opposite direction) (d',a') by \theta to end up at (d'',a'') and then lower (or raise) this point by c to end up at (d'',a''-c). That point would be the reflection point.

    ok, you need to find the line that is perpendicular to that line you mention, but which also contains the point a,b, ....see math book.

    then you would need to find the point of the intersection of those two lines,....solve the linear system.

    then you would need to make that intersection point the average, or midpoint, of the two reflected points, but solve that equation for the unknown reflected point, say a', b'.

    i hope that helps!
    This is a bad approach. VERY messy. Try it.
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  6. #6
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    I am not familiar with how to rotate points through an angle, i know it can be done through some matrix transformation but I needs t set aside some time to read about them can you recommend some ( preferably free online ) reading material.

    I managed to get this done in the end using vectors, with a reasonable neat answer.
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  7. #7
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    Quote Originally Posted by bobak View Post
    I am not familiar with how to rotate points through an angle, i know it can be done through some matrix transformation but I needs t set aside some time to read about them can you recommend some ( preferably free online ) reading material.
    Let (x_0,y_0) be a point different from the origin. Let (x_1,y_1) be its image under a rotation by \theta degrees (counterclockwise). Draw a line l_0 connecting the origin with (x_0,y_0) and draw l_1 connecting the origin with (x_1,y_1). Let \alpha be the angle l_0 makes with the x-axis and let \beta be the angle l_1 makes with the x-axis. Then (assuming \alpha > \beta) \alpha = \theta + \beta. This means \sin \alpha = \sin (\theta + \beta) = \sin \beta \cos \theta + \cos \beta \sin \theta. Let r = \sqrt{x_0^2+y_0^2} be the distance from the point to the origin, but then r = \sqrt{x_1^2+y_1^2} because the distance of this rotated point maintains its distance from the origin. Multiply by r to get r\sin \alpha = r\sin \beta \cos \theta + r\cos \beta \sin \theta thus y_1 = y_0\cos \theta + x_0 \sin \theta. And if you do the same with \cos \alpha = \cos (\theta + \beta ) = \cos \beta \cos \theta - \sin \beta \sin \theta we would get x_1 = x_0 \cos \theta - y_0 \sin \theta. We can write this as a matrix:
    \left( \begin{array}{c}x_1\\y_1 \end{array}\right) = \left( \begin{array}{cc}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{array} \right)\left( \begin{array}{c} x_0\\y_0 \end{array} \right)
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