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: ![\[x+2y=3\]](http://latex.artofproblemsolving.com/9/c/3/9c3cc3de86dae8a6d15929d4bfc1cee1c4cc3bf8.png)
![\[x+3y=5\]](http://latex.artofproblemsolving.com/5/2/b/52bbca35323da434d502013ab88b64b0f41c48c8.png)
Use elimination: ![\[y = 2\]](http://latex.artofproblemsolving.com/2/4/d/24d6268c8f2a2c9fee56a2090135a47741eb6487.png)
Plug this into one of the previous lines. ![\[x+4 = 3 \Rightarrow x=-1\]](http://latex.artofproblemsolving.com/c/d/d/cddaecac8d8386d5b0db0ac822dd0d446fb57314.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
