line symmetry

given the point A(u,v) and the line y=ax+b, 

find the point B(p,q) which is symmetrical with respect to A.



the middle point between A and B is

((u+p)/2, (v+q)/2), and the point satisfies the line y=ax+b, so we get


(v+q)/2 = a*(u+p)/2 + b ⇔ ap - q = v - au - 2b   ・・・(Σ)


|AC| = |BC| then we get


root(u^2+(v-b)^2) = root(p^2+(q-b)^2)  ・・・(Λ)


solving (Σ) and (Λ) for p and q


(subject to a^2+1!=0,)  we get


p =

     -a^2 u - 2ab + 2av + u




q =

      a^2 v + 2au + 2b - v





when y=3x+2, A(1,7), find the point B(p,q).


p =


    --------------------------------  = 11/5



q =


     --------------------------- = 33/5




the formulas of p and q are complex, so 

you do not have to remember them.


all you have to do is to find the two equations,

v+7        u+1

---- = 3 *------ + 2  ⇔ v = 3u

  2            2




u^2 +(v-2)^2 = 26, and you solve them.

other solutions

finding the line through B(p,q) that is vertical to the line y=ax+b,

that is q = -1/a (p-u) + b (a=!0).


solving this and (Σ) above allows you to find the point (p,q).




if you can find the two lines m and n, you can get the answer by solving them.

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