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

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Baby you're not that kind
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All lines with equation <math>ax+by=c</math> such that <math>a,b,c</math> form an arithmetic progression pass through a common point. What are the coordinates of that point?
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<math>\textbf{(A) } (-1,2)
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\qquad\textbf{(B) } (0,1)
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\qquad\textbf{(C) } (1,-2)
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\qquad\textbf{(D) } (1,0)
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\qquad\textbf{(E) } (1,2)</math>
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==Solution==
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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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Use elimination: <cmath>y = 2</cmath> Plug this into one of the previous lines. <cmath>x+4 = 3 \Rightarrow x=-1</cmath> Thus the common point is <math>\boxed{A) (-1,2)}</math>

Revision as of 14:49, 14 February 2019

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

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 $\boxed{A) (-1,2)}$