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

I keep getting VERY messy answers to this question, anyone want to give me a few pointers?

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- Feb 6th 2008, 03:13 PMbobakReflection in line
Find the cooridiantes of the reflection of the point $\displaystyle (a , b)$ in the line $\displaystyle y = mx + c$.

I keep getting VERY messy answers to this question, anyone want to give me a few pointers? - Feb 6th 2008, 07:58 PMpellikkan
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.

- Feb 7th 2008, 06:39 AMbobak
- Feb 7th 2008, 07:18 AMpellikkan
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! - Feb 7th 2008, 07:19 AMThePerfectHacker
I assume this approach will work. It eliminates a lot of mess.

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

Quote:

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!

- Feb 11th 2008, 08:20 AMbobak
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. - Feb 11th 2008, 10:07 AMThePerfectHacker
Let $\displaystyle (x_0,y_0)$ be a point different from the origin. Let $\displaystyle (x_1,y_1)$ be its image under a rotation by $\displaystyle \theta$ degrees (counterclockwise). Draw a line $\displaystyle l_0$ connecting the origin with $\displaystyle (x_0,y_0)$ and draw $\displaystyle l_1$ connecting the origin with $\displaystyle (x_1,y_1)$. Let $\displaystyle \alpha$ be the angle $\displaystyle l_0$ makes with the x-axis and let $\displaystyle \beta$ be the angle $\displaystyle l_1$ makes with the x-axis. Then (assuming $\displaystyle \alpha > \beta$) $\displaystyle \alpha = \theta + \beta$. This means $\displaystyle \sin \alpha = \sin (\theta + \beta) = \sin \beta \cos \theta + \cos \beta \sin \theta$. Let $\displaystyle r = \sqrt{x_0^2+y_0^2}$ be the distance from the point to the origin, but then $\displaystyle r = \sqrt{x_1^2+y_1^2}$ because the distance of this rotated point maintains its distance from the origin. Multiply by $\displaystyle r$ to get $\displaystyle r\sin \alpha = r\sin \beta \cos \theta + r\cos \beta \sin \theta$ thus $\displaystyle y_1 = y_0\cos \theta + x_0 \sin \theta$. And if you do the same with $\displaystyle \cos \alpha = \cos (\theta + \beta ) = \cos \beta \cos \theta - \sin \beta \sin \theta$ we would get $\displaystyle x_1 = x_0 \cos \theta - y_0 \sin \theta$. We can write this as a matrix:

$\displaystyle \left( \begin{array}{c}x_1\\y_1 \end{array}\right) = \left( \begin{array}{cc}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{array} \right)\left( \begin{array}{c} x_0\\y_0 \end{array} \right)$