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

(Solution 3)
(Solution 3)
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===Solution 3===
 
===Solution 3===
 
We use process of elimination. <math>\textbf{B}</math> doesn't necessarily work because <math>b = c</math> isn't always true.  <math>\textbf{C, D, E}</math> also doesn't necessarily work because the x-value is <math>1</math>, but the y-value is an integer. So by process of elimination, <math>\boxed{\textbf{(A) } (-1, 2)}</math> is our answer.    ~Baolan
 
We use process of elimination. <math>\textbf{B}</math> doesn't necessarily work because <math>b = c</math> isn't always true.  <math>\textbf{C, D, E}</math> also doesn't necessarily work because the x-value is <math>1</math>, but the y-value is an integer. So by process of elimination, <math>\boxed{\textbf{(A) } (-1, 2)}</math> is our answer.    ~Baolan
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==Solution 4==
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We know that in <math>ax + by = c</math>, <math>a</math>, <math>b</math>, and <math>c</math> are in an arithmetic progression. We can simplify any arithmetic progression to be <math>0</math>, <math>1</math>, <math>2</math>, and <math>-1</math>, <math>0</math>, <math>1</math>.
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For example, the progression <math>2</math>, <math>4</math>, <math>6</math> can be rewritten as <math>0</math>, <math>2</math>, <math>4</math> by going back by one value. We can then divide all 3 numbers by 2 which gives us <math>0</math>, <math>1</math>, <math>2</math>.
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Now, we substitute <math>a</math>, <math>b</math>, and <math>c</math> with <math>0</math>, <math>1</math>, <math>2</math>, and <math>-1</math>, <math>0</math>, <math>1</math> respectively. This gives us
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<math>y = 2</math> and <math>-x = 1</math> which can be written as <math>x = -1</math>. The only point of intersection is <math>(-1,2)</math>. So, our answer is
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<math>\boxed{\textbf{(A) } (-1, 2)}</math>.
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~Starshooter11
  
 
==See Also==
 
==See Also==

Revision as of 15:49, 8 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

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
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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