2019 AMC 10A Problems/Problem 7: Difference between revisions
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==Solution 1== | ==Solution 1== | ||
The two lines are <math>y= 2x-2</math> and <math>y = x | The two lines are <math>y=2x-2</math> and <math>y = \frac{x}{2}+1</math>, which intersect the third line at <math>(4,6)</math> and <math>(6,4)</math>. So we have an isosceles triangle with base <math>2\sqrt{2}</math> and height <math>3\sqrt{2} \implies \boxed{\textbf{(C)}6}</math>. | ||
==Solution 2== | ==Solution 2== | ||
Revision as of 18:51, 9 February 2019
- The following problem is from both the 2019 AMC 10A #7 and 2019 AMC 12A #5, so both problems redirect to this page.
Problem
Two lines with slopes 
and 
intersect at 
. What is the area of the triangle enclosed by these two lines and the line 
Solution 1
The two lines are 
and 
, which intersect the third line at 
and 
. So we have an isosceles triangle with base 
and height 
.
Solution 2
Like in Solution 1, let's first calculate the slope-intercept form of all three lines:
becomes 
so b=1,
becomes 
so c=-2, and
becomes 
.
(
and .)
Now we find the intersections between each of the lines with , which are and . Applying the Shoelace Theorem, we can find that the solution is 
See Also
| 2019 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 | ||
| 2019 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 6 |
| 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 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
