Difference between revisions of "2019 AMC 10B Problems/Problem 4"
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− | + | We use process of elimination. <math>\textbf{B}</math> doesn't necessarily work because <math>b = c</math> isn't always true. <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. ~Baolan | |
==See Also== | ==See Also== |
Revision as of 13:55, 2 February 2020
Problem
All lines with equation such that
form an arithmetic progression pass through a common point. What are the coordinates of that point?
Solution
Solution 1
If all lines satisfy the condition, then we can just plug in values for ,
, and
that form an arithmetic progression. Let's use
,
,
, and
,
,
. Then the two lines we get are:
Use elimination to deduce
and plug this into one of the previous line equations. We get
Thus the common point is
.
~IronicNinja
Solution 2
We know that ,
, and
form 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 (regardless of the values of
and
), we must have
and
, so
and the common point is
.
Solution 3
We use process of elimination. doesn't necessarily work because
isn't always true.
also doesn't necessarily work because the x-value is
, but the y-value is an integer. So by process of elimination,
is our answer. ~Baolan
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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