SOLUTION: Point B is the reflection of point A=(2,-1) over the line y=1. Point C is the reflection of point B over the line x + 2y = a. If C=(-5,8), what is a?

Question 1210142: Point B is the reflection of point A=(2,-1) over the line y=1. Point C is the reflection of point B over the line x + 2y = a. If C=(-5,8), what is a?
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let $A = (2, -1)$.
Since $B$ is the reflection of $A$ over the line $y=1$, the $x$-coordinate of $B$ is the same as the $x$-coordinate of $A$, and the $y$-coordinate of $B$ is such that the midpoint of $AB$ lies on the line $y=1$.
Let $B = (x_B, y_B)$.
The midpoint of $AB$ is $\left(\frac{2+x_B}{2}, \frac{-1+y_B}{2}\right)$.
Since this midpoint lies on $y=1$, we have $\frac{-1+y_B}{2} = 1$, so $-1+y_B = 2$, and $y_B = 3$.
Thus, $B = (2, 3)$.
Now, $C = (-5, 8)$ is the reflection of $B = (2, 3)$ over the line $x + 2y = a$.
Let the midpoint of $BC$ be $M = \left(\frac{2+(-5)}{2}, \frac{3+8}{2}\right) = \left(-\frac{3}{2}, \frac{11}{2}\right)$.
Since $M$ lies on the line $x + 2y = a$, we have
$$-\frac{3}{2} + 2\left(\frac{11}{2}\right) = a$$
$$-\frac{3}{2} + 11 = a$$
$$-\frac{3}{2} + \frac{22}{2} = a$$
$$a = \frac{19}{2}$$
The line $BC$ is perpendicular to the line $x + 2y = a$.
The slope of $x + 2y = a$ is $-\frac{1}{2}$.
The slope of $BC$ is $\frac{8-3}{-5-2} = \frac{5}{-7} = -\frac{5}{7}$.
Since the slopes are not negative reciprocals, our midpoint calculation is sufficient.
Therefore, $a = \frac{19}{2}$.
Final Answer: The final answer is $\boxed{\frac{19}{2}}$

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