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

The two lines are $y=2x-2$ and $y = \frac{x}{2}+1$ , which intersect the third line at $(4,6)$ and $(6,4)$ . So we have an isosceles triangle with base $2\sqrt{2}$ and height $3\sqrt{2} \implies \boxed{\textbf{(C) }6}$ .

Solution 2

Like in Solution 1, let's first calculate the slope-intercept form of all three lines: $(x,y)=(2,2)$ and $y=x/2 + b$ implies $2=2/2 +b=1+b$ so $b=1$ , while $y=2x + c$ implies $2= 2*2+c=4+c$ so $c=-2$ . Also, $x+y=10$ implies $y=-x+10$ . Thus the lines are $y=x/2 +1, y=2x-2,$ and $y=-x+10$ . Now we find the intersections between each of the lines with $y=-x+10$ , which are $(6,4)$ and $(4,6)$ . Applying the Shoelace Theorem, we can find that the solution is $6 \implies \boxed{\textbf{(C) }6}.$

Solution 3

Like the other solutions, solve the systems to see that the triangles two other points are at $(4, 6)$ and $(6, 4)$ . The apply Heron's Formula. The semi-perimeter will be $s = \sqrt{2} + \sqrt{20}$ . The area then reduces nicely to a difference of squares, making it $6 \implies \boxed{\textbf{(C) }6}.$

Solution 4

Using the methods above to find the coordinates, we get $(4, 6)$ , $(6, 4)$ , and $(2,2)$ . Using the Shoelace Theorem, we find that it is equal to $(4\cdot 2)-(2\cdot 6)+(2\cdot 4)-(6\cdot 2)+(6 \cdot 6)-(4\cdot 4) = 12$ . Because the Shoelace Theorem tells us to find the half of that sum, we get $\boxed{\text{6 (C)}}$

--Claire (clara32356)

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.

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