Triangle $ABC$ is isosceles with point $A$ at the point $(2, 7)$, with point $B$ at $(-2, 0)$ and with point $C$ at $(3, -1).$ Triangle $ABC$ is reflected over $\overline{BC}$ to form $\triangle A'BC$. Triangle $A'BC$ is reflected over $\overline{A'C}$ to form $\triangle A'B'C$. Triangle $A'B'C$ is reflected over $\overline{B'C}$ to form $\triangle A''B'C$. What is the slope of $\overline{A''A}$? Express your answer as a common fraction.
After all of the reflections, we get a figure that is symmetric with the line of symmetry being $A'C$. The midpoint of $\overline{BC}$ is $(0.5,-0.5)$, which is also the midpoint of $\overline{AA'}$. Thus we find that $A'$ has the coordinates $(-1,-8)$. The line of symmetry $\overline{A'C}$ has the slope of $\frac74$.
Since the triangles are isosceles and $A''B'C$ is a reflection of $ABC$ across the $\overline{A'C}$, $\overline{AA''}$ is perpendicular to $\overline{A'C}$. Therefore, the slope of $\overline{AA''}$ is the negative of the reciprocal of the slope of $\overline{A'C}$, which is $\boxed{-\frac47}$.
Well, let's get our reflection game on!
First, we'll reflect triangle $ABC$ over line $\overline{BC}$, which is essentially like folding it in half. This gives us $\triangle A'BC$.
Next, we'll reflect $\triangle A'BC$ over line $\overline{A'C}$, which is like folding it in half again. This forms $\triangle A'B'C$.
Finally, we'll reflect $\triangle A'B'C$ over line $\overline{B'C}$, folding it one more time. This gives us $\triangle A''B'C$.
Now, we need to find the slope of $\overline{A''A}$. Well, the slope of a line is simply the change in $y$ coordinates divided by the change in $x$ coordinates.
The $y$ coordinate of $A''$ is $-1$, and the $y$ coordinate of $A$ is $7$. So the change in $y$ is $7 - (-1) = 8$.
Similarly, the $x$ coordinate of $A''$ is $3$, and the $x$ coordinate of $A$ is $2$. So the change in $x$ is $2 - 3 = -1$.
Therefore, the slope of $\overline{A''A}$ is $\frac{8}{-1}$, which simplifies to $\boxed{-8}$.
To find the slope of $\overline{A''A}$, we need to find the coordinates of points $A''$ and $A$ first.
1. Reflection over $\overline{BC}$:
To reflect a point across a line, we use the formula $P' = (2q - x, 2s - y)$, where $(x, y)$ is the original point and $P'$ is the reflection of the point across the line $2qx + 2sy + t = 0$. In this case, the line is $\overline{BC}$, which passes through $(-2, 0)$ and $(3, -1)$. So the equation of the line is $2x - 5y - 6 = 0$. Applying the formula, the coordinates of $A'$ can be found as follows:
$$A' = (2(-5)(2) - 2, 2(-5)(7) - 0) = (-24, -70)$$
2. Reflection over $\overline{A'C}$:
Using the formula again, the coordinates of $A''$ can be found:
$$A'' = (2(-5)(-24) - (-24), 2(-5)(-70) - (-70)) = (-72, -200)$$
3. Now we can find the slope of $\overline{A''A}$:
The slope of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$. Applying this formula, we get:
$$m = \frac{-70 - (-200)}{-24 - (-72)} = \frac{130}{-48} = \boxed{-\frac{65}{24}}$$
So the slope of $\overline{A''A}$ is $-\frac{65}{24}$.
To find the slope of $\overline{A''A}$, we need to first find the coordinates of points $A''$ and $A$.
Since point $A$ is reflected over $\overline{BC}$ to form $\triangle A'BC$, the $x$-coordinate of point $A'$ will be the same as the $x$-coordinate of $A$. However, the $y$-coordinate of point $A'$ will be the reflection of the $y$-coordinate of $A$ about the line $\overline{BC}$.
The equation of the line $\overline{BC}$ can be found by finding the slope and y-intercept. The slope of a line passing through points $B(-2,0)$ and $C(3,-1)$ can be found using the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$. Substituting the coordinates of $B$ and $C$, we get $m = \frac{-1 - 0}{3 - (-2)} = \frac{-1}{5}$.
Since the line passes through $B(-2,0)$, the equation of the line can be written as $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept. Substituting the slope and the coordinates of $B$, we have $0 = \frac{-1}{5} \cdot (-2) + b$, which gives $b = \frac{2}{5}$.
Therefore, the equation of line $\overline{BC}$ is $y = \frac{-1}{5}x + \frac{2}{5}$.
To find the coordinates of $A'$, we need to reflect point $A(2,7)$ over line $\overline{BC}$. The reflection of a point $(x, y)$ over a line with equation $y = mx + b$ can be found using the formula $(x', y') = \left(\frac{x - m(y - b)}{1 + m^2}, \frac{y - m(x - b)}{1 + m^2}\right)$. Substituting the coordinates of point $A$ and the equation of line $\overline{BC}$, we get
\begin{align*}
x' &= \frac{2 - \frac{-1}{5}(7 - \frac{2}{5})}{1 + (\frac{-1}{5})^2}\\
&= \frac{2 + \frac{7}{5} - \frac{2}{5}}{1 + \frac{1}{25}}\\
&= \frac{\frac{25}{5} + \frac{7}{5} - \frac{2}{5}}{\frac{26}{5}}\\
&= \frac{\frac{30}{5}}{\frac{26}{5}}\\
&= \frac{30}{26}\\
&= \frac{15}{13},
\end{align*}
and
\begin{align*}
y' &= \frac{7 - \frac{-1}{5}(2 - \frac{2}{5})}{1 + (\frac{-1}{5})^2}\\
&= \frac{7 + \frac{2}{5} - \frac{2}{5}}{1 + \frac{1}{25}}\\
&= \frac{\frac{35}{5} + \frac{2}{5} - \frac{2}{5}}{\frac{26}{5}}\\
&= \frac{\frac{35}{5}}{\frac{26}{5}}\\
&= \frac{35}{26},
\end{align*}
Hence, point $A'$ is at $\left(\frac{15}{13}, \frac{35}{26}\right)$.
Now, we need to find the coordinates of $A''$. Since point $A'$ is reflected over $\overline{A'C}$ to form $\triangle A'B'C$, the $y$-coordinate of point $A''$ will be the same as the $y$-coordinate of $A'$. However, the $x$-coordinate of point $A''$ will be the reflection of the $x$-coordinate of $A'$ about the line $\overline{B'C}$.
Using the same method as above, we can find the equation of line $\overline{B'C}$ to be $y = \frac{1}{5}x - \frac{1}{5}$.
To find the coordinates of $A''$, we need to reflect point $A'$ over line $\overline{B'C}$. Substituting the coordinates of point $A'$ and the equation of line $\overline{B'C}$ into the formula, we get
\begin{align*}
x'' &= \frac{\frac{15}{13} - \frac{1}{5}\left(\frac{35}{26}\right) - \left(\frac{1}{5}\right)}{1 + \left(\frac{1}{5}\right)^2}\\
&= \frac{\frac{75}{65} - \frac{35}{130} - \frac{13}{65}}{\frac{26}{25}}\\
&= \frac{\frac{75 - 35 - 13}{65}}{\frac{26}{25}}\\
&= \frac{\frac{27}{65}}{\frac{26}{25}}\\
&= \frac{27}{65} \cdot \frac{25}{26}\\
&= \frac{27}{26},
\end{align*}
and
\begin{align*}
y'' &= \frac{\frac{35}{26} - \frac{1}{5}\left(\frac{15}{13} - \frac{15}{13}\right)}{1 + \left(\frac{1}{5}\right)^2}\\
&= \frac{\frac{35}{26}}{\frac{26}{25}}\\
&= \frac{35}{26} \cdot \frac{25}{26}\\
&= \frac{35}{26},
\end{align*}
Hence, point $A''$ is at $\left(\frac{27}{26}, \frac{35}{26}\right)$.
Now, we have the coordinates of points $A''$ and $A$. We can calculate the slope of $\overline{A''A}$ using the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$. Substituting the coordinates of $A''$ and $A$, we get
\begin{align*}
m &= \frac{\frac{35}{26} - \frac{35}{26}}{\frac{27}{26} - \frac{15}{13}}\\
&= \frac{0}{\frac{351}{338}}\\
&= \boxed{0}.
\end{align*}
Therefore, the slope of $\overline{A''A}$ is $0$.