Question

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.

Answer 1

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 $\boxed{-\frac47}$.

Answer 2

1/47