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

--------------------------------

a^2+1

q =

a^2 v + 2au + 2b - v

---------------------------

a^2+1

question

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

p =

-2×3×2+1-3^2×1+2×3×7

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

3^2+1

q =

2×2+2×3×1-7+3^2×7

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

3^2+1

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

and

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).

or

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