Difference between revisions of "2019 AMC 10B Problems/Problem 4"
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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> | 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> | ||
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> | 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> | ||
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==Solution 2== | ==Solution 2== |
Revision as of 11:13, 15 February 2019
Contents
Problem
All lines with equation such that form an arithmetic progression pass through a common point. What are the coordinates of that point?
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: Use elimination: Plug this into one of the previous lines. Thus the common point is
Solution 2
We know that are an arithmetic progression, so if the common difference is we can say Now we have , and expanding gives Factoring gives . Since this must always be true, we know that and , so and the common point is .
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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