Difference between revisions of "2019 AMC 10B Problems/Problem 4"

Revision as of 11:49, 17 June 2020

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

https://youtu.be/kB_dR5H7Pzw

~savannahsolver

@@ Line 35: / Line 35: @@
 <math>\boxed{\textbf{(A) } (-1, 2)}</math>.
 ~Starshooter11
+==Video Solution==
+https://youtu.be/kB_dR5H7Pzw
+~savannahsolver
 ==See Also==

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

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