2006 AMC 10B Problem 12

1. The point $(1, 2)$ lies on both lines. So $x = 1$ and $y = 2$ make both equations true.
2. First line, $x = \frac{1}{4}y + a$. Put in $x = 1$ and $y = 2$:
$$1 = \frac{1}{4}\cdot 2 + a = \frac{1}{2} + a, \quad\text{so}\quad a = \frac{1}{2}.$$
3. Second line, $y = \frac{1}{4}x + b$. Put in $x = 1$ and $y = 2$:
$$2 = \frac{1}{4}\cdot 1 + b = \frac{1}{4} + b, \quad\text{so}\quad b = \frac{7}{4}.$$
4. Add them, using quarters: $a + b = \frac{2}{4} + \frac{7}{4} = \frac{9}{4}$.

The answer is $\boxed{\textbf{(E)}\ \frac{9}{4}}$.

A point is on a line exactly when its coordinates make the line's equation true. So a known point lets you find a missing number in the equation. Here the meeting point is known, so there is nothing to solve for $x$ and $y$; just substitute into each equation separately.

An intersection point satisfies both equations, so put $x = 1$ and $y = 2$ into each.

First line: $1 = \frac{1}{4}(2) + a$, so $a = 1 - \frac{1}{2} = \frac{1}{2}$.

Second line: $2 = \frac{1}{4}(1) + b$, so $b = 2 - \frac{1}{4} = \frac{7}{4}$.

Therefore $a + b = \frac{1}{2} + \frac{7}{4} = \frac{2}{4} + \frac{7}{4} = \frac{9}{4}$.

The answer is $\boxed{\textbf{(E)}\ \frac{9}{4}}$.