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

Revision as of 14:49, 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

@@ Line 23: / Line 23: @@
 ===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
+==Solution 4==
+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>.
+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>.
+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
+<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
+<math>\boxed{\textbf{(A) } (-1, 2)}</math>.
+~Starshooter11
 ==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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