How can I find the coordinates of a point reflected over a line that may not necessarily be any of the axis?

Example Question:

If P is a reflection (image) of point (3, -3) in the line $2y = x+1$, find the coordinates of Point P.

I know the answer is $(-1,5)$ by drawing a graph but other than that, I cannot provide any prior workings because I don't know how to start...

  • $\begingroup$ Do you know the formula for finding the foot of the perpendicular from a given point to a given line? $\endgroup$ – Tejas Nov 9 '14 at 12:32
  • $\begingroup$ No, I'm afraid I don't. Mind enlightening me? $\endgroup$ – Happytreat Nov 9 '14 at 12:36
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    $\begingroup$ Sure. Though I am afraid I don't know how to prove it. $\endgroup$ – Tejas Nov 9 '14 at 12:38
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    $\begingroup$ Assume the point to be $Q(h,k)$, the mid point of $PQ$ must satisfy the equation of line and its slope must be perpendicular to it. You will get two equations, solve it and you can get $Q$'s coordinates. Hope this helps, although I am very late. :) $\endgroup$ – dark32 Dec 17 '16 at 9:51

The formula for finding the foot of the perpendicular from a point $(x_1,y_1)$ to the line $ax+by+c=0$ is given by: $$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-(ax_1+by_1+c)}{a^2+b^2}$$

For finding the image of the point in the same line, we just multiply the rightmost term by 2.

So, the image of the point $(x_1,y_1)$ in the line $ax_1+by_1+c=0$ is given by: $$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-2(ax_1+by_1+c)}{a^2+b^2}$$

The image of the point is at the same distance from the line as the point itself is from the line. So, we have to multiply it by 2. That's what I think.

Here's the proof from my book: enter image description here

enter image description here

Please excuse the image size.. :P

  • $\begingroup$ Can you please tell me the name of the pictured book? $\endgroup$ – Ian Mar 21 at 17:28
  • $\begingroup$ You can find it here. I believe there's a newer edition. $\endgroup$ – Tejas Mar 23 at 1:02
  • $\begingroup$ Thank you for your feedback. I was anticipating an ebook version. I searched for it without success. $\endgroup$ – Ian Mar 28 at 20:56

Find line perpendicular to $ 2y=x+1$ which passes through $(3, -3)$


Find intersection point of those two lines ( Foot of perpendicular )


Use Mid-Point Formula (special case of Section formula) to get required point (Foot of perpendicular is mid-point of reflection and original point)



In general, the reflection of the point $(p, q)$ in the line $ay + bx + c = 0$ is


Proof here (no complex numbers needed!).


So in this case, your point is $(p,q)=(3,-3)$ and your line is $2y-x-1=0$ (rewritten in the form $ay+bx+c=0$). So just plug in the numbers to get $(-1,5)$ as desired.

Or you can use the spreadsheet titled "Reflection of Point in Line" here.

(I was looking for a general solution but couldn't seem to find any online, so I thought I might as well share my work here.)


I like to use complex numbers for this. Let's first do it when the symmetry axis passes through the origin. Let $\alpha$ be the angle it forms with the $x$-axis and set $u=\cos\alpha+i\sin\alpha$; then you get the symmetric point of $z\in\mathbb{C}$ by first rotating by $-\alpha$, then conjugating and then rotating again by $\alpha$: the first rotation makes the axis of symmetry coincide with the $x$-axis, so the reflection is just conjugation.

Thus you do (remembering that $u^{-1}=\bar{u}$) $$ z\mapsto \bar{u}z\mapsto \overline{\bar{u}z}\mapsto u\overline{\bar{u}z}=u^2\bar{z} $$ If we go back to real numbers, we get $$ x+iy\mapsto (x\cos2\alpha+y\sin2\alpha)+i(x\sin2\alpha-y\cos2\alpha) $$ so the equation is $$ (x,y)\mapsto \left(\frac{(1-m^2)x+2my}{1+m^2},\frac{2mx-(1-m^2)y}{1+m^2} \right) $$ if the equation of the line is $y=mx$, because $m=\tan\alpha$ and so $$ \cos2\alpha=\frac{1-m^2}{1+m^2},\quad\sin2\alpha=\frac{2m}{1+m^2}. $$

If the line has equation $y=mx+q$ one just adds first a translation by $-iq$ and last a translation by $iq$: $$ z\mapsto z-iq\mapsto \bar{u}(z-iq)\mapsto \overline{\bar{u}(z-iq)}\mapsto u\overline{\bar{u}(z-iq})\mapsto u\overline{\bar{u}(z-iq})+iq $$ and, finally, $$ z\mapsto u^2\bar{z}+iq(1+u^2) $$ Note that $1+u^2=1+\cos2\alpha+i\sin2\alpha=\dfrac{2}{1+m^2}+i\dfrac{2m}{1+m^2}$ so the equation is $$ (x,y)\mapsto \left(\frac{(1-m^2)x+2my-2mq}{1+m^2},\frac{2mx-(1-m^2)y+2q}{1+m^2} \right) $$ In the present case $m=1/2$ and $q=1/2$, so, just by substituting, $$ (3,-3)\mapsto (-1,5) $$

Of course this is overkill if you just want to know the image of one point, but I believe the method well illustrates the power of complex numbers.

  • $\begingroup$ Whew that's pretty darn neat $\endgroup$ – Vrisk Nov 2 '17 at 14:09
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    $\begingroup$ @Vrisk Last week I assigned the task to my students in the “mathematics teaching” course, telling them to “think to this complex problem”. $\endgroup$ – egreg Nov 2 '17 at 14:16

Here is one way: there are many others.

Let the desired image point be $(a,b)$. Then you want two conditions:

1) The midpoint of the line segment between the given point and the image point is on the given line. This gives us $$2\frac{-3+b}2= \frac{3+a}2+1$$ 2) The line containing the given point and the image point is perpendicular to the given line. This means their slopes are negative reciprocals, giving us $$\frac{-3-b}{3-a}=-\frac{1}{1/2}$$

These are two equations in two unknowns, letting us find $a$ and $b$.


Let me me add another one for passers by :

We can get the perpendicular distance from the point tk the line by :

$\frac{ax+by+c}{\sqrt{a^2 + b^2}} $

where a, b, c are coefficients of the line and x and y is the coordinates of your given point.

Here $ a=1,b=1,c=-2;x=3,y=-3$

we find this length to be $\frac{10}{\sqrt{5}}$

The direction vector of the line is $(1,\frac{1}{2})$ By observation we find the vector perpendicular this representing the perpendicular from point to the line must have direction

$(- \frac{1}{2}, 1)$

On normalisation -

$(- \frac{1}{2}, 1) \frac{2}{\sqrt{5}}$

so the vector from the point to the foot of the perpendicular is just this times the distance between ie

$(- \frac{1}{2}, 1)\times \frac{2}{\sqrt{5}} \times 2 \sqrt{5}$( found above.)

Thus our vector is $(-2,4)$. We want to have this twice fhe length for reflection. so $(-4,8)$. Now finally ae obtain this point by vectorially adding our starting point (draw it!)


which isssss

$(-1,5) $


Similar answer to @Vrisk, but a bit faster

Consider the line $L=Ax+By+C=0$ and find image of point $(u,v)$ assuming $Au + Bv + C \neq 0 $

If we extend the space which the equation exists into $R^3$ , we will find that the equation $Ax+By+C=0$ denotes a plane with a unit normal vector as : $$ \hat{n} = \frac{<A,B>}{\sqrt{A^2 +B^2} } $$

Now, consider the unsigned perpendicular distance of this point from the line/plane :

$$ d = \frac{Au +Bv +C}{\sqrt{A^2 +B^2}}$$

Now, depending on sign of the above quantity we can see which part of the plane/ line which the point is (see here).

If, from our point we were translate in twice the direction of unit normal scaled up by the distance from line to point, we would reach the image point. Hence, the image coordinates$(I)$ are given by the vector:

$$I =<u,v>- \frac{2(Au+Bv+c)}{A^2 +B^2} <A,B>$$

Or more concisely:

$$I= <u,v> -2d \hat{n}$$

Here is a useful diagram: enter image description here


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