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 .
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 .
Super fast solution
Simply transform 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: (4, 6)(6, 4)s = \sqrt{2} + \sqrt{20}\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}$.
==Solution 5== Like in other solutions, we find that the three points of intersection are$ (Error compiling LaTeX. Unknown error_msg)(2, 2)(4, 6)(6, 4)64\frac62 + 4 - 1 = \boxed{\textbf{(C) }6}$.
==Solution 6== Like in other solutions, we find the three points of intersection. Label these$ (Error compiling LaTeX. Unknown error_msg)A (2, 2)B (4, 6)C (6, 4)AB = AC = 2\sqrt5BC = 2\sqrt2\sin A = \sqrt{1 - \cos^2 A} = \frac35\frac12 bc \sin A = \frac12 \cdot 2\sqrt5 \cdot 2\sqrt5 \cdot \frac35 = \boxed{\textbf{(C) }6}$.
==Solution 7== Like in other solutions, we find that the three points of intersection are$ (Error compiling LaTeX. Unknown error_msg)(2, 2)(4, 6)(6, 4)$. The area of the triangle is half the absolute value of the determinant of the matrix determined by these points. <cmath> \frac12\begin{Vmatrix} 2&2&1\\ 4&6&1\\ 6&4&1\\ \end{Vmatrix} = \frac12|-12| = \frac12 \cdot 12 = \boxed{\textbf{(C) }6}</cmath>
==Solution 8==
Like in other solutions, we find the three points of intersection. Label these$ (Error compiling LaTeX. Unknown error_msg)A (2, 2)B (4, 6)C (6, 4)\overrightarrow{AB} = \langle 2, 4 \rangle\overrightarrow{AC} = \langle 4, 2 \rangle$. The area of the triangle is half the magnitude of the cross product of these two vectors.
<cmath>
\frac12\begin{Vmatrix}
i&j&k\\
2&4&0\\
4&2&0\\
\end{Vmatrix} = \frac12|-12k| = \frac12 \cdot 12 = \boxed{\textbf{(C) }6}</cmath>
==Solution 9== Like in other solutions, we find that the three points of intersection are$ (Error compiling LaTeX. Unknown error_msg)(2, 2)(4, 6)(6, 4)2\sqrt22\sqrt5bl\frac{b\sqrt{4l^2-b^2}}{4}b = 2\sqrt2l = 2\sqrt5$, <cmath>\frac{2\sqrt2 \cdot \sqrt{80-8}}{4} = \frac{\sqrt{576}}{4} = \frac{24}{4} = \boxed{\textbf{(C) }6}</cmath>
==Solution 10 (Trig) == Like in other solutions, we find the three points of intersection. Label these$ (Error compiling LaTeX. Unknown error_msg)A (2, 2)B (4, 6)C (6, 4)AB = AC = 2\sqrt5BC = 2\sqrt2\sin A = \sqrt{1 - \cos^2 A} = \frac35\frac{abc}{4R} = \frac{2\sqrt2 \cdot 2\sqrt5^2}{4 \cdot \frac{5\sqrt2}{3}} = \boxed{\textbf{(C) }6}$.
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.
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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