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

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(Solution)
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\qquad\textbf{(E) } (1,2)</math>
 
\qquad\textbf{(E) } (1,2)</math>
  
==Solution==
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==Solution 1==
  
 
If all lines satisfy the equation, then we can just plug in values for a, b, and c that form an arithmetic progression. Let's do a=1, b=2, c=3 and a=1, b=3, and c=5. Then the two lines we get are: <cmath>x+2y=3</cmath> <cmath>x+3y=5</cmath>
 
If all lines satisfy the equation, then we can just plug in values for a, b, and c that form an arithmetic progression. Let's do a=1, b=2, c=3 and a=1, b=3, and c=5. Then the two lines we get are: <cmath>x+2y=3</cmath> <cmath>x+3y=5</cmath>
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iron
 
iron
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==Solution 2==
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We know that <math>a,b,c</math> are 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, we know that <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{A) (-1,2)}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 18:26, 14 February 2019

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 equation, then we can just plug in values for a, b, and c that form an arithmetic progression. Let's do a=1, b=2, c=3 and a=1, b=3, and c=5. Then the two lines we get are: \[x+2y=3\] \[x+3y=5\] Use elimination: \[y = 2\] Plug this into one of the previous lines. \[x+4 = 3 \Rightarrow x=-1\] Thus the common point is $\boxed{A) (-1,2)}$

iron

Solution 2

We know that $a,b,c$ are 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, we know that $x+y-1 = 0$ and $y-2 = 0$, so $x,y = -1, 2,$ and the common point is $\boxed{A) (-1,2)}$.

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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