2019 AMC 10B Problem 4

If every such line passes through one point, then any two of them locate it. Take two easy progressions.

• $(a, b, c) = (1, 1, 1)$: the line $x + y = 1$.
• $(a, b, c) = (1, 2, 3)$: the line $x + 2y = 3$.

Subtracting the first equation from the second gives $y = 2$, and then $x = -1$. So the point is $(-1, 2)$.

Confirm it works in general: an arithmetic progression satisfies $b = \frac{a + c}{2}$, i.e. $c = 2b - a$. At $(-1, 2)$ the left side is $a(-1) + b(2) = 2b - a = c$, so the point lies on every line in the family.

The answer is $\boxed{\textbf{(A)}\ (-1,2)}$.

A family of lines that all share a point is determined by any two members, so picking the simplest two and intersecting them is the fastest route. The verification step shows why: the arithmetic-progression condition $c = 2b - a$ is itself a linear relation in $a$ and $b$, and its coefficients $(-1, 2)$ are the coordinates.

Test the choices in the general condition: substituting $(x, y)$ into $ax + by = c$ with $c = 2b - a$ requires $ax + by = -a + 2b$ for all $a, b$, forcing $x = -1$, $y = 2$ directly.

• Solving the general condition a - 2b + c = 0 and reading off (1,-2) with the signs backwards, since the line is ax + by = c, not ax + by + c = 0.
• Testing only one line (say $x + y = 1$) and picking a choice that lies on it, such as $(0, 1)$ or $(1, 0)$.