Find the reflection of the point $(4,-13)$ in the line $5x+y+6=0$

Question

Find The image(or reflection) of the point $(4,-13)$ in the line $5x+y+6=0$

Method 1 $$ y+13=\frac{1}{5}(x-4)\implies x-5y-69=0\quad\&\quad 5x+y+6=0\implies (3/2,-27/2)\\ (3/2,-27/2)=(\frac{x+4}{2},\frac{y-13}{2})\implies(x,y)=(-1,-14) $$

Method 2

$m=\tan\theta=-5$

Ref$(\theta)$=$\begin{bmatrix} \cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta) \end{bmatrix}$ $$ \cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{1-25}{1+25}=\frac{-24}{26}=\frac{-12}{13}\\ \sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{-10}{26}=\frac{-5}{13}\\ Ref(\theta)\begin{bmatrix}4\\-13\end{bmatrix}=\begin{bmatrix} \cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta) \end{bmatrix}\begin{bmatrix}4\\-13\end{bmatrix}=\begin{bmatrix} \dfrac{-12}{13} & \dfrac{-5}{13} \\ \dfrac{-5}{13} & \dfrac{12}{13} \end{bmatrix}\begin{bmatrix}4\\-13\end{bmatrix}\\ =\frac{1}{13}\begin{bmatrix}-48+65\\-20-156\end{bmatrix}=\frac{1}{13}\begin{bmatrix}17\\-176\end{bmatrix} $$

Why am I not getting the required solution in Method two using matrix method ?

Thanx @ganeshie8 for the remarks, so in that case how do I find the operator for reflection of a point over the line not passing through the origin ?

Answer 1

As ganeshie8 suggested, your matrix formula is not working because the line does not pass through the origin.

When you translate everything up by $6$ units, the line now passes through the origin and you can continue as follows:

$$\begin{bmatrix} \dfrac{-12}{13} & \dfrac{-5}{13} \\ \dfrac{-5}{13} & \dfrac{12}{13} \end{bmatrix}\begin{bmatrix}4\\-7\end{bmatrix}\\$$

$$=\frac{1}{13} \begin{bmatrix} -48+35 \\ -20-84 \end{bmatrix}\\$$

$$= (-1, -8)$$

And now translate down by $6$ units to find that the original coordinate is at $(-1,-14)$.

Answer 2

Translating up by $6$ units again, we find that the angle between the $x$-axis and $(4,-7)$ is $-\tan^{-1} \frac{7}{4}$. In addition, the angle between the line and the $x$-axis is $-\tan^{-1} 5$. So the reflected point must have an angle of:

$$\theta =-\tan^{-1} 5 -\tan^{-1} 5 + \tan^{-1} \frac{7}{4}$$

In addition, the distance between the origin and the original point is the same as the distance between the origin and the reflected point (since the reflected angles are equal). This gives:

$$r = \sqrt{4^2 + 7^2}$$

Now, the reflected point is at $(r \cos \theta, r \sin \theta) = (-1,-8)$. Translating down by $6$ units, we again get $(-1, -14)$.

Answer 3

The simplest (and shortest) way is to do some affine geometry:

Find first the projection of the point $A(4,-13)$ onto the line $5x+y+6=0$. As a directing vector for this projection is $\vec n (5,1)$, you have a parametric equation of the line of projection: $$\overrightarrow{OM}=A+t\mkern 1.5mu\vec n, $$it suffices to find $t$ so the point $M$ satisfies the equation $5x+y+6=0$. Then, the reflection of $A$ is the point $$A'=A+2t\mkern 1.5mu \vec n.$$

Answer 4

Method 3. Translate the origin to the point $(\color{red}{\frac32,-\frac{27}{2}})$ and find the coordinates of the point ${\color{blue}{4\choose -13}}$ in the new system: $${x'\choose y'}={x\color{red}{-\frac32}\choose y\color{red}{+\frac{27}{2}}}={\color{blue}4-\frac32\choose \color{blue}{-13}+\frac{27}{2}}={\frac52 \choose \frac{1}{2}}$$ Rotate it by $180^\circ$: $${\begin{pmatrix}-1&0\\ 0&-1\end{pmatrix}}{\frac52\choose \frac12}={-\frac52\choose -\frac12}$$ Now translate it back: $${x\choose y}={-\frac52+\frac32\choose -\frac12-\frac{27}{2}}={-1\choose -14}.$$