Difference between revisions of "2019 AMC 10A Problems/Problem 7"
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<math>\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}</math> | <math>\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}</math> | ||
| − | ==Solution== | + | ==Solution 1== |
The two lines are <math>y= 2x-2</math> and <math>y = 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}</math> <math>\implies \boxed{(C) 6}</math>. | The two lines are <math>y= 2x-2</math> and <math>y = 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}</math> <math>\implies \boxed{(C) 6}</math>. | ||
| + | ==Solution 2== | ||
| + | Like in Solution 1, let's first calculate the slope-intercept form of all three lines: | ||
| + | <math>(x,y)=(2,2)</math> | ||
| + | <math>y=x/2 + b</math> becomes <math>2=2/2 +b=1+b</math> so b=1, | ||
| + | <math>y=2x + c</math> becomes <math>2= 2*2+c=4+c</math> so c=-2, and | ||
| + | <math>x+y=10</math> becomes <math>y=-x+10</math>. | ||
| + | (<math>y=x/2 +1, y=2x-2,</math> and <math>y=-x+10</math>.) | ||
| + | Now we find the intersections between each of the lines with <math>y=-x+10</math>, which are <math>(6,4)</math> and <math>(4,6)</math>. Applying the Shoelace Theorem, we can find that the solution is <math>6 \implies \boxed{C}.</math> | ||
==See Also== | ==See Also== | ||
Revision as of 19:05, 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.
Contents
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
