2019 AMC 10A Problems/Problem 7
- 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
Let's first work out the slope-intercept form of all three lines:
and
implies
so
, while
implies
so
. Also,
implies
. Thus the lines are
and
.
Now we find the intersection points between each of the lines with
, which are
and
. Using the distance formula and then the Pythagorean Theorem, we see that we have an isosceles triangle with base
and height
, whose area is \boxed{\textbf{(C) }6}$.
==Solution 2== Like in Solution 1, we determine the coordinates of the three vertices of the triangle. Now, using the [[Shoelace Theorem]], we can directly find that the area is$ (Error compiling LaTeX. Unknown error_msg)\boxed{\textbf{(C) }6}$.
==Solution 3==
Like in the other solutions, solve the systems of equations to see that the triangle's two other vertices are at$ (Error compiling LaTeX. Unknown error_msg)(4, 6)(6, 4)
s = \sqrt{2} + \sqrt{20}
\implies \boxed{\textbf{(C) }6}$.
==Solution 4==
Like in the other solutions, we find, either using algebra or simply by drawing the lines on squared paper, that the three points of intersection are$ (Error compiling LaTeX. Unknown error_msg)(2, 2)(4, 6)
(6, 4)
(2, 2)
(2, 6)
(6, 6)
(6, 2)
16-4-2-4=\boxed{\textbf{(C) }6}$.
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.