A point (a,b) is reflected over the line Ax + By + C = 0. What is the new point?

I think I have yet another way to derive the result.





http://answers.yahoo.com/question/index?...





y = -A/B x - C/B





Move C/B units up in the vertical direction:





y = -A/B y





(a,b) becomes (a, b+C/B) = (a,c)





the angle between the line y = -A/B y and y = x is arctan(-A/B) - pi/4





rotate the point (a,c) about the above angle and reflect across y = x by switching the y- and x-coordinates:





a' = a cos(pi/4 - arctan(-A/B)) - c sin(pi/4 - arctan(-A/B)) = c''





c' = c cos(pi/4 - arctan(-A/B)) + a sin(pi/4 - arctan(-A/B)) = a''





rotate the point (a'',c'') about the angle arctan(-A/B) - pi/4:





a''' = a'' cos(pi/4 - arctan(-A/B)) + c'' sin(pi/4 - arctan(-A/B))





c''' = c'' cos(pi/4 - arctan(-A/B)) - a'' sin(pi/4 - arctan(-A/B))





Finally, translate the point A/B downward in the vertical direction:


(a''', c''' - A/B).





I wonder if this can be simplified to


x' = ((b²- a²)x - 2a(by + c)) / (a²+b²)


y' = ((a² - b²)y - 2b(ax + c)) / (a²+b²)

A point (a,b) is reflected over the line Ax + By + C = 0. What is the new point?
If you are interested in another methodology, we can use vectors.


I am going to give a numerical example to demonstrate the method





Say we want the reflection of A(1, 2) about 2x + 3y = 6





The line through (1,2) perpendicular to 2x + 3y = 6 is





(1 + 2t, 2 + 3t) and it intersects 2x + 3y = 6 at





2 + 4t + 6 + 9t = 6


13t = 2


t = -2/13


point of intersection M( 9/13, 20/13)





M is the midpoint of AA'


OM = 1/2 OA + 1/2 OA'





OA' = 2OM - OA = (18/13, 40/13) - ( 1,2) = (5/13, 14/13)





In agreement with the result obtained by substituting into formula given bellow.





It looks messy with letters so I will leave it at that.
Reply:This is an offshoot from this question, rather than an answer to it. It relates to your comment about proving that a combination of two reflections about two intersecting lines is a rotation about the point of intersection through double the angle required to rotate the first line onto the second. I take that as proven, and address this problem:


A rotation about point A through angle x, followed by a rotation about point B through angle y, is equivalent to a rotation through angle x+y. But where is the centre of this single rotation?


There is a very simple construction for locating this centre, obtained as follows:


Through the point A, draw a line k inclined at angle -x/2 to the line AB. That is, rotating k about A through angle x/2 brings it into coincidence with AB. Then


M(AB) M(k) is equivalent to rotation about A through angle x, where M(u) denotes reflection in line u and operations are carried out from right to left.


Through B, draw a line s inclined at angle y/2 to AB. Then


M(s) M(AB) is equivalent to rotation about B through angle y.


Hence rotation about A through angle x followed by rotation about B through angle y is equivalent to


M(s) M(AB) M(AB) M(k)


= M(s) M(k) since reflection is self-inverse.


Angle properties show that s is inclined to k at an angle


x/2 + y/2 [Either use vertically opposite angles and exterior angle of a triangle, or draw a line thru intersection point, parallel to AB, and use corresponding angles and alternate angles]


therefore the combination of rotations is equivalent to a rotation through angle x+y about the point of intersection of k and s.
Reply:If You wish to get it in the most straightforward way, don't forget that the analytic geometry is the most successful try to "algebrize" the geometry in the history of science. Some even say that in "analytic geometry" the adjective contains 99% of the essence, the noun - 1%. So forget about rotations, translations, angles and think algebraically, not geometrically.


Let P'(x', y') is the given point, P"(x", y") is the symmetric point with respect to the line L with the equation:


L: Ax + By + C = 0 /A, B, C = const, at least one of A, B not 0/


Remind that (A, B) is a non-zero vector, normal (perpendicular) to L. The distance from P' to L is:


|(Ax' + By' + C)/(A² + B²)|


and the ORIENTED DISTANCE is the same expression without the absolute value. P' and P" are symmetric, hence their oriented distances are sign-negated, since the points are in opposite semi-planes with respect to L:


(Ax' + By' + C)/(A² + B²) = -(Ax" + By" + C)/(A² + B²)


and that yields


1) Ax" + By" = -Ax' - By' - 2C


The symmetry requires P'P" to be perpendicular to L, hence the vectors P'P"(x" - x', y" - y') and (A, B) to be collinear, so


their coordinates must be proportional, what leads to:


2) -Bx" + Ay" = -Bx' + Ay'


Finally all You need is to solve the system 1) - 2) for x", y". The Cramer rule /the determinant A² + B² ≠ 0!/ yields exactly both expressions in Your question /in my notations/:


x" = ((B² - A²)x' - 2A(By' + C)) / (A² + B²)


y" = ((A² - B²)y' - 2B(Ax' + C)) / (A² + B²)
Reply:♠ the equation of the line L is c*x +s*y =d, where


c=A/√(A^2 +B^2), s=B/√(A^2 +B^2), d=-C/√(A^2 +B^2) in your terms;


♦ c %26amp; s are components of unit vector i’ normal to L, while unit vector


j’=[k×i’]=(-s,c) is directed parallel to L, and


d is bias (displacement) of L from origin (0,0);


drawn a picture?


♣ with above told vector (a,b)=d*i’+p*j’+g*i’, where


p %26amp; g are measured with respect to L;


thence ((a,b)·j’)=p, and ((a,b)·i’)=d+g;


♥ the mirror point in question is


(a’,b’)= d*i’+p*j’-g*i’;


(a’,b’)= [2d -((a,b)·i’)]*i’ +((a,b)·j’)*j’=


=(2d -ac -bs)*(c,s) +(-as +bc)*(-s,c) =


=(2dc -acc -bsc +a*ss -bcs, 2ds -acs -bss -asc +bcc)=


=(2dc -a(c^2-s^2) -2bsc, 2ds -2asc +b(c^2-s^2)) =


= (-2AC –a*(A^2-B^2) -2bAB,


-2BC +b*(A^2-B^2) -2aAB) /(A^2+B^2);


▬ to your additional details; looks bizarre for me, I can’t follow it sorry; check numerically if you understand the stuff;
Reply:starwhitedwarf has the most straight forward method.





Using her example. Reflect the point A(1, 2) across the line





2x + 3y = 6





The perpendicular line thru A(1, 2) that intersects the given line is:





r(t) = %26lt;1, 2%26gt; + t%26lt;2, 3%26gt; = %26lt;1 + 2t, 2 + 3t%26gt;





As she notes, plug this into the formula of the given line and solve for t.





2x + 3y = 6


2(1 + 2t + 3(2 + 3t) = 6


t = -2/13





To find the reflection point simply double t.


t = 2(-2/13) = -4/13





x' = 1 + 2t = 1 - 2(-4/13) = 1 - 8/13 = 5/13


y' = 2 + 3t = 2 + 3(-4/13) = 2 - 12/13 = 14/13





The reflected point is A'(5/13, 14/13).





As you can see, this agrees with starwhitedwarf's answer.

palm