Reflecting a triangle over a line $y=2x+1$ 
A triangle with vertices at $A(-2,2), B(-8,2)$ and $C(-8,-1)$ is reflected about the line $y=2x+1$. Express the coordinates of the reflection of A as an ordered pair.
(A) $(-2,0)$  (B) $(0,-2)$  (C) $(-2,8)$  (D) $(2,-2)$  (E) $(2,0)$.

I know that in the reflection about line $y=x+m$ you do $(x,y)\mapsto(y-m,x+m)$. I don't even know if this is the right one because in this case m will have to be $x+1$ and for me it lead to no where.
Someone please help me if you can.
 A: $E$. Because, the direction vector of the line is $\vec u=\langle1,2\rangle$ and $\vec{AE}=\vec E-\vec A=\langle 4,-2\rangle$ and check: $\vec u\cdot\vec{AE}=0.$
(Also, the distance of $A$ to the line is $h=\frac{|2(-2)+1-2|}{\sqrt{2^2+1^2}}=\sqrt{5}$ and $AE=\sqrt{4^2+(-2)^2}=2\sqrt{5}$ and $\frac{AE}{h}=2$, as required.)
A: OP's formula is for a special line $y=x+m$ and it will not work here because the given line is $2x-y+1=0$.
In general, the image $(x,y)$ of the point $(h,k)$ in the line $ax+by+c=0$ is given by
$$\frac{x-h}{a}=\frac{y-k}{b}=-2 \frac{ah+bk+c}{a^2+b^2}.$$
Here $a=2,b=-1, c=1$
For the image of $A(-2,0)$, we get
$$\frac{x+2}{2}=\frac{y-0}{-1}=-2\frac{-4-2+1}{5}\implies A'(2,2)$$
For the image of $B(0,-2)$, we have
$$\frac{x-0}{2}=\frac{y+2}{-1}=-2\frac{0+2+1}{5}\implies B'(-12/5,-4/5)$$
For the image of $C(-2,8)$, we get
$$\frac{x+2}{2}=\frac{y-8}{-1}=-2\frac{-4-8+1}{5}\implies C'(-34/5,18/5)$$
for $D(2.-2)$, we get
$$\frac{x-2}{2}=\frac{y+2}{-1}=-2\frac{4+2+1}{5}\implies D'(-18/5,14/5)$$
For $E(2,0)$, we gget
$$\frac{x-2}{2}=\frac{y-0}{-1}=-2\frac{4-1+1}{5}\implies E'(-6/5,8/5)$$
A: Hint:
The image $(x,y)$ of the point $(h,k)$ in the line $ax+by+c=0$ is given by
$$\frac{x-h}{a}=\frac{y-k}{b}=-2 \frac{ah+bk+c}{a^2+b^2}.$$
