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

Revision as of 12:50, 2 February 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

Use quick process of elimination. $\textbf{B}$ doesn't necessarily work because $b!=c$ . $\textbf{C, D, E}$ 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.

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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All AMC 10 Problems and Solutions

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@@ Line 19: / Line 19: @@
 ===Solution 2===
 We know that <math>a</math>, <math>b</math>, and <math>c</math> form an arithmetic progression, so if the common difference is <math>d</math>, we can say <math>a,b,c = a, a+d, a+2d.</math> Now we have <math>ax+ (a+d)y = a+2d</math>, and expanding gives <math>ax + ay + dy = a + 2d.</math> Factoring gives <math>a(x+y-1)+d(y-2) = 0</math>. Since this must always be true (regardless of the values of <math>x</math> and <math>y</math>), we must have <math>x+y-1 = 0</math> and <math>y-2 = 0</math>, so <math>x,y = -1, 2,</math> and the common point is <math>\boxed{\textbf{(A) } (-1,2)}</math>.
+===Solution 3===
+Use quick process of elimination. <math>\textbf{B}</math> doesn't necessarily work because <math>b!=c</math>. <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.
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