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2006 AMC 10B Problem 12

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2006 AMC 10B Problem 12AMC 10 Step by Step

Name: Date: September 13, 2026

Coordinate Geometry · difficulty 2 of 5 · about 2 min · position P11-15

The lines $x=\frac{1}{4}y+a$ and $y=\frac{1}{4}x+b$ intersect at the point $(1,2)$ . What is $a+b$ ?

A)
$0$
B)
$\frac{3}{4}$
C)
$1$
D)
$2$
E)
$\frac{9}{4}$

Answer

Key insight

The intersection point lies on both lines, so substituting x = 1, y = 2 into each equation solves for a and b immediately.

Solution

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

Why this works

Unknown constants in a line's equation are found by feeding in a known point; the intersection point is simply a point known to be on both lines. There is no system to solve, only two substitutions. Notice the shortcut: $a + b = (x - \tfrac{y}{4}) + (y - \tfrac{x}{4}) = \tfrac{3}{4}(x + y) = \tfrac{3}{4} \cdot 3$ .

The trap

Swapping the roles of x and y between the two equations, or trying to solve the system for x and y when they are already given.

Common mistakes

Swapping the roles of x and y between the two equations, or trying to solve the system for x and y when they are already given.
Arithmetic slips with the fractions, such as $1 - \tfrac{1}{4}$ for $a$ , which yields $\tfrac{5}{2}$ or $2$ .

Techniques

Set up the equation/formula and compute; no special trick needed · Test small/specific values or special cases to find or verify the answer