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How to find the coordinates of the reflection of the point $(1, 0)$ in the line $y = mx$?

I tried, but I can't think of any way to start this problem. Any help is greatly appreciated.

Eris Tyenns
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  • have you tried drawing it for a fixed m? – Riquelme Aug 25 '19 at 14:42
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    What do you mean by 'reflection'? How is that defined? – Cornman Aug 25 '19 at 14:43
  • Yes, I tried drawing it for a fixed m, but I can't find the coordinates. – Eris Tyenns Aug 25 '19 at 14:45
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    Can you find coordinates of the point on the line closest to $(1,0)$? – J. W. Tanner Aug 25 '19 at 14:46
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    You can do this with straightforward algebra, or with linear algebra since it is a linear transformation. If you go with the first option, why not first try y=2x and reflect your x-intercept. Perpendicular slope is -1/2 , set up line y-0=-1/2(x-1),find point of intersection with y=2x, etc... then generalize from there. – imranfat Aug 25 '19 at 14:46
  • It is an exercise I'm currently trying to solve, I wrote it exactly as it is written in the exam question, so I am not sure what exactly they mean by reflection.@Cornman – Eris Tyenns Aug 25 '19 at 14:46
  • I don't know how to find them, since I don't know the value of m. How could I find them? @J.W.Tanner – Eris Tyenns Aug 25 '19 at 14:48
  • Try finding the coordinate of the closest point on the line to (1,0), and then reflect across the point. – Gabe Aug 25 '19 at 14:50
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    Ok, so if I generalise, the slope is -1/m and the line perpendicular is y=(-x+1)/m. The point of interaction of the line y=mx and the perpendicular line is ($\frac{1}{1+m^2}$,$\frac{m}{1+m^2}$). What can you do from there to find the reflection?@imranfat – Eris Tyenns Aug 25 '19 at 14:51
  • The point of reflection is on the other side of y=mx. So count from (1,0) to your point of intersection you know those distances. – imranfat Aug 25 '19 at 14:55
  • keep $m$ a variable; the distance from $(1,0)$ to the point of reflection is twice the distance and along the same direction from $(1,0)$ to the point of intersection – J. W. Tanner Aug 25 '19 at 15:01
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    @ErisTyenns If $A = (0, 1)$ and $B = (\frac 1{1+m^2}, \frac m {1 + m^2})$ are known, and $C$ is the reflection point, you know $A - B = B - C$, solve for $C$. – DanielV Aug 25 '19 at 15:44

3 Answers3

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Hint:

Let's do some geometry: if $\varphi$ is the polar angle of the given line, and $\theta$ the polar angle of the reflection , say $M$, we know that the vector $\overrightarrow{OM}$ will be a unit vector, so it has coordinates $(\cos\theta,\sin\theta)$.

On the other hand, the points $M$ and $(1,0)$ are the reflection of each other w.r.t. the given line if and only if $\theta=2\varphi$. So by the duplication formulæ in trigonometry, we have $$\tan\theta=\frac{2\tan \varphi}{1-\tan^2\varphi}=\frac{2m}{1-m^2}\qquad (m\ne\pm1).$$ Can you continue?

Bernard
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Observe that the reflection point is on the line that is normal to $ y=mx$ and passes through (1,0), i.e.

$$y= -\frac{1}{m}(x-1)$$

which intersects with the line $y=mx$ at

$$ \left( \frac{1}{m^2+1}, \frac{m}{m^2+1} \right)$$

Let (a,b) be the reflection point, then the above intercept is the midpoint of (1,0) and (a,b). So, simply take the average,

$$ \frac{1+a}{2} = \frac{1}{m^2+1}, \space \frac{b}{2}=\frac{m}{m^2+1} $$

which solves the reflection point at,

$$\left( \frac{1-m^2}{1+m^2}, \space \frac{2m}{1+m^2} \right)$$

Quanto
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Consider a line perpendicular to $y=mx$ through $(1,0):$

$y=-(1/m)(x-1)$;

The reflected point $(x',y')$ has distance $1$ to origin (why?).

Hence:

$(x')^2+ (y')^2=$

$ (x')^2+(1/m^2)(x'-1)^2=1$;

$m^2(x')^2+(x'-1)^2-m^2=0$;

$(1+m^2)(x')^2-2x' +(1-m^2)=0;$

$x'_{1,2}=$

$\dfrac{2\pm\sqrt{4-4(1+m^2)(1-m^2)}}{2(1+m^2)}$;

$x'_{1,2}=\dfrac {1\pm m^2}{1+m^2}$;

$x'_1=1$;

$x'_2= \dfrac{1-m^2}{1+m^2}$;

Reflected point:

$(x'_2, -(1/m)(x'_2-1)).$

Peter Szilas
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