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

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

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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

Page

Toolbox

Search

2019 AMC 10B Problems/Problem 4

Contents

Problem

Solution 1

Solution 2

See Also