# 2019 AMC 10B Problems/Problem 4

## Problem

All lines with equation $ax+by=c$ such that $a,b,c$ form an arithmetic progression pass through a common point. What are the coordinates of that point? $\textbf{(A) } (-1,2) \qquad\textbf{(B) } (0,1) \qquad\textbf{(C) } (1,-2) \qquad\textbf{(D) } (1,0) \qquad\textbf{(E) } (1,2)$

## Solution

### Solution 1

If all lines satisfy the condition, then we can just plug in values for $a$, $b$, and $c$ that form an arithmetic progression. Let's use $a=1$, $b=2$, $c=3$, and $a=1$, $b=3$, $c=5$. Then the two lines we get are: $$x+2y=3$$ $$x+3y=5$$ Use elimination to deduce $$y = 2$$ and plug this into one of the previous line equations. We get $$x+4 = 3 \Rightarrow x=-1$$ Thus the common point is $\boxed{\textbf{(A) } (-1,2)}$.

~IronicNinja

### Solution 2

We know that $a$, $b$, and $c$ form an arithmetic progression, so if the common difference is $d$, we can say $a,b,c = a, a+d, a+2d.$ Now we have $ax+ (a+d)y = a+2d$, and expanding gives $ax + ay + dy = a + 2d.$ Factoring gives $a(x+y-1)+d(y-2) = 0$. Since this must always be true (regardless of the values of $x$ and $y$), we must have $x+y-1 = 0$ and $y-2 = 0$, so $x,y = -1, 2,$ and the common point is $\boxed{\textbf{(A) } (-1,2)}$.

### Solution 3

We use process of elimination. $\textbf{B}$ doesn't necessarily work because $b = c$ isn't always true. $\textbf{C, D, E}$ also doesn't necessarily work because the x-value is $1$, but the y-value is an integer. So by process of elimination, $\boxed{\textbf{(A) } (-1, 2)}$ is our answer. ~Baolan

## Solution 4

We know that in $ax + by = c$, $a$, $b$, and $c$ are in an arithmetic progression. We can simplify any arithmetic progression to be $0$, $1$, $2$, and $-1$, $0$, $1$.

For example, the progression $2$, $4$, $6$ can be rewritten as $0$, $2$, $4$ by going back by one value. We can then divide all 3 numbers by 2 which gives us $0$, $1$, $2$.

Now, we substitute $a$, $b$, and $c$ with $0$, $1$, $2$, and $-1$, $0$, $1$ respectively. This gives us $y = 2$ and $-x = 1$ which can be written as $x = -1$. The only point of intersection is $(-1,2)$. So, our answer is $\boxed{\textbf{(A) } (-1, 2)}$. ~Starshooter11

## Video Solution

~savannahsolver

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 