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.
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
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 .
Solution 2
Like in Solution 1, we determine the coordinates of the three vertices of the triangle. The coordinates that we get are: . Now, using the Shoelace Theorem, we can directly find that the area is .
BEST solution
Simply translate to the origin, which would change to . Now the three points are , , . Now, because it is at the origin, we can easily take the half the discriminant: ~N828335
Solution 3
Like in the other solutions, solve the systems of equations to see that the triangle's two other vertices are at and . Then apply Heron's Formula: the semi-perimeter will be , so the area reduces nicely to a difference of squares, making it .
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 , , and . We can now draw the bounding square with vertices , , and , and deduce that the triangle's area is .
Solution 5
Like in other solutions, we find that the three points of intersection are , , and . Using graph paper, we can see that this triangle has boundary lattice points and interior lattice points. By Pick's Theorem, the area is .
Solution 6
Like in other solutions, we find the three points of intersection. Label these , , and . By the Pythagorean Theorem, and . By the Law of Cosines, Therefore, , so the area is .
Solution 7
Like in other solutions, we find that the three points of intersection are , , and . The area of the triangle is half the absolute value of the determinant of the matrix determined by these points.
Solution 8
Like in other solutions, we find the three points of intersection. Label these , , and . Then vectors and . The area of the triangle is half the magnitude of the cross product of these two vectors.
Solution 9
Like in other solutions, we find that the three points of intersection are , , and . By the Pythagorean theorem, this is an isosceles triangle with base and equal length . The area of an isosceles triangle with base and equal length is . Plugging in and ,
Solution 10 (Trig)
Like in other solutions, we find the three points of intersection. Label these , , and . By the Pythagorean Theorem, and . By the Law of Cosines, Therefore, . By the extended Law of Sines, Then the area is .
Solution 11
The area of a triangle formed by three lines, is the absolute value of Plugging in the three lines, the area is the absolute value of Source: Orrick, Michael L. “THE AREA OF A TRIANGLE FORMED BY THREE LINES.” Pi Mu Epsilon Journal, vol. 7, no. 5, 1981, pp. 294–298. JSTOR, www.jstor.org/stable/24336991.
Video Solution
~savannahsolver
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 |
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