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 $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$

$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$

Solution 1

Let's first work out the slope-intercept form of all three lines: $(x,y)=(2,2)$ and $y=\frac{x}{2} + b$ implies $2=\frac{2}{2} +b=1+b$ so $b=1$ , while $y=2x + c$ implies $2= 2 \cdot 2+c=4+c$ so $c=-2$ . Also, $x+y=10$ implies $y=-x+10$ . Thus the lines are $y=\frac{x}{2} +1, y=2x-2,$ and $y=-x+10$ . Now we find the intersection points between each of the lines with $y=-x+10$ , which are $(6,4)$ and $(4,6)$ . Using the distance formula and then the Pythagorean Theorem, we see that we have an isosceles triangle with base $2\sqrt{2}$ and height $3\sqrt{2}$ , 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) $and$ (6, 4) $. Then apply Heron's Formula: the semi-perimeter will be$ s = \sqrt{2} + \sqrt{20} $, so the area reduces nicely to a difference of squares, making it$ \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) $and$ (6, 4) $. We can now draw the bounding square with vertices$ (2, 2) $,$ (2, 6) $,$ (6, 6) $and$ (6, 2) $, and deduce that the triangle's area is$ 16-4-2-4=\boxed{\textbf{(C) }6}$.

2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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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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2019 AMC 10A Problems/Problem 7

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Solution 1

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