2019 AMC 10B Problem 4

The problem promises that all these lines meet at one point. So two of them are enough to find it. Pick the simplest arithmetic progressions.

1. $a, b, c = 1, 1, 1$ gives the line $x + y = 1$.
2. $a, b, c = 1, 2, 3$ gives the line $x + 2y = 3$.
3. Subtract the first equation from the second: $y = 2$.
4. Put $y = 2$ into $x + y = 1$: $x = -1$.

So the point should be $(-1, 2)$.

Check it with a third line, just to be safe. Take $a, b, c = 2, 5, 8$ (common difference $3$): $2(-1) + 5(2) = -2 + 10 = 8$. Yes, $(-1, 2)$ is on $2x + 5y = 8$ too.

Why it always works: in an arithmetic progression the middle term is the average of the other two, so $c = 2b - a$. Plugging in $x = -1$, $y = 2$ gives $a(-1) + b(2) = 2b - a$, which is exactly $c$.

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

When a problem says "all of these pass through a common point," you do not need the general case to find it. Two simple members of the family pin the point down, and a third member is a quick check. The general reason is that the progression rule $c = 2b - a$ looks exactly like the line equation with $x = -1$ and $y = 2$.

Test the answer choices in a single line of the family, such as $x + 2y = 3$ (from $1, 2, 3$). The correct point must satisfy it.

• $(-1, 2)$: $-1 + 4 = 3$. Fits.
• $(0, 1)$: $0 + 2 = 2$. No.
• $(1, -2)$: $1 - 4 = -3$. No.
• $(1, 0)$: $1 + 0 = 1$. No.
• $(1, 2)$: $1 + 4 = 5$. No.

Only (A) survives.

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