Reflection in a line

**alexgeek** · September 6th 2010, 08:44 AM

Couldn't find much useful on the net for this so I'll ask here.
If you're given a question such as find the reflection of a point (12, 1) in the line y=7x, how would you go about this?
I think there's a way using some trig to measure how far the point is away from the line and then using that some how but not really sure how to go about it or if it will work.
Thanks

**yeKciM** · September 6th 2010, 09:02 AM

Originally Posted by alexgeek

Couldn't find much useful on the net for this so I'll ask here.
If you're given a question such as find the reflection of a point (12, 1) in the line y=7x, how would you go about this?
I think there's a way using some trig to measure how far the point is away from the line and then using that some how but not really sure how to go about it or if it will work.
Thanks

to find symmetrical point on the line, you should first do distance of that point on the line ....

if your line is :

$\displaystyle \frac {x-x_1}{m} = \frac {y-y_1}{n}=\frac {z-z_1}{p}$

and you have some point at $A_2 (x_2 ,y_2 , z_2)$

to find distance between line $l$ and point $A_2$ you can do like this

(or you can use formula, but let's do it like this so you understand what and how is done)

Let's first mark one point at the line as $A_1 (x_1, y_1, z_1)$ and on start of the line vector $\vec {q} = (m,n,q)$ we put that point $A_1$ (now look at the picture shown, than continue)

distance $d$ from the point $A_2$ and line $l$ is equal to ratio of surface of the parallelogram constructed on vectors $\vec{q}$ and $\vec {A_1 A_2}$ and intensity of the vector $\vec {q}$

as shown :

$\displaystyle d= \frac {|\vec{q} \times \vec {A_1 A_2 }|}{|\vec {q}| }$

or in scalar form

$\displaystyle d = \frac {\sqrt{ \begin{vmatrix} y_1-y_2 &z_1-z_2 \\ n & p \end{vmatrix}^2 + \begin{vmatrix} x_1-x_2 &z_1-z_2 \\ m & p \end{vmatrix}^2 +\begin{vmatrix} x_1-x_2 &y_1-y_2 \\ m & n \end{vmatrix}^2 }}{\sqrt{m^2+n^2+p^2}}$

can you figure the rest ? i'll go now

bye

**alexgeek** · September 6th 2010, 09:02 AM

Think I found what I need using the matrix $\begin{pmatrix} \cos 2 \theta & \sin 2 \theta \\ \sin 2 \theta & - \cos 2 \theta \end{pmatrix}$ to reflect a point in the line $y=(tan \theta)x$

-edit-
Sorry seemed to post at the same time as you, I'm just reading your post now.

-edit 2-
Ok, brains stopped working now.. too much maths for one day. Have a look at that again tomorrow ha
Thanks!

**Soroban** · September 6th 2010, 11:00 AM

Hello, alexgeek!

$\text{Find the reflection of a point }(12, 1)\text{ in the line }y\,=\,7x$

Code:

      |
Q     |
o     |         / y = 7x
   *  |        /
      *       /
      |  *  M/
      |     o
      |    /   *
      |   /       *
      |  /           *
      | /               o (12,1)
      |/                P 
- - - + - - - - - - - - - - - - 
      |

We want a line perpendicular to $y \,=\,7x.$ .[1]

This line has slope $\text{-}\frac{1}{7}$ and contains (12,1).

. . Its equation is: . $y -1 \:=\:\text{-}\frac{1}{7}(x-12) \quad\Rightarrow\quad y \:=\:\text{-}\frac{1}{7}x + \frac{19}{7}$ .[2]

$M$ is the intersection of [1] and [2].

. . $7x \:=\:\text{-}\frac{1}{7}x + \frac{19}{7} \quad\Rightarrow\quad 7x + \frac{1}{7}x \:=\:\frac{19}{7} \quad\Rightarrow\quad \frac{50}{7}x \:=\:\frac{19}{7}$

. . $x \:=\:\frac{19}{50} \quad\Rightarrow\quad y \:=\:\frac{133}{50}$

Point $M$ is $\left(\dfrac{19}{50},\:\dfrac{133}{50}\right)$

Let point $Q$ be $(x,y)$
Note that $M$ is the midpoint of $PQ.$

So we have: . $\begin{Bmatrix}\dfrac{x+12}{2} &=& \dfrac{19}{50} \\ \\[-3mm] \dfrac{y+1}{2} &=& \dfrac{133}{50} \end{Bmatrix}$

Solve the two equations: . $x \,=\,\text{-}\dfrac{281}{25},\;y \,=\,\dfrac{108}{25}$

Therefore: . $Q\left(\text{-}\dfrac{281}{25},\;\dfrac{108}{25}\right)$

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