2019 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 7 Easy

Two lines with slopes \frac{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? (2019 AMC 10A Problem, Question#7)

  • A.

    4

  • B.

    4\sqrt{2}

  • C.

    6

  • D.

    8

  • E.

    6\sqrt{2}

Answer:C

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\cdot2+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 6.

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 6.

Like in the other solutions, solve the systems of equations to see that the triangle's two other vertices are at (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 6.

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 (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=6.

Like in other solutions, we find that the three points of intersection are (2,2), (4,6), and (6,4). Using graph paper, we can see that this triangle has 6 boundary lattice points and 4 interior lattice points. By Pick's Theorem, the area is \frac{6}{2}+4-1=6.

Like in other solutions, we find the three points of intersection. Label these A(2,2), B(4,6), and C(6,4). By the Pythagorean Theorem, AB=AC=2\sqrt{5} and BC=2\sqrt{2}. By the Law of Cosines, \cos A=\frac{(2\sqrt{2})^2+(2\sqrt{5})^2-(2\sqrt{2})^2}{2\cdot2\sqrt{5}\cdot2\sqrt{5}}=\frac{20+20-8}{40}=\frac{32}{40}=\frac{4}{5} Therefore, \sin A=\sqrt{1-\cos^2 A}=\frac{3}{5}, so the area is \frac{1}{2}bc\sin A=\frac{1}{2}\cdot2\sqrt{5}\cdot2\sqrt{5}\cdot\frac{3}{5}=6.

Like in other solutions, we find that the three points of intersection are (2,2), (4,6), and (6,4). The area of the triangle is half the absolute value of the determinant of the matrix determined by these points. \dfrac{1}{2}\left| \left| \begin{array}{l} {2\ \ 2\ \ 1}\\{4\ \ 6\ \ 1}\\{6\ \ 4\ \ 1} \end{array}\right|\right|=\dfrac{1}{2}\left| -12\right|=\dfrac{1}{2}\cdot12=6.

Like in other solutions, we find the three points of intersection. Label these A(2,2), B(4,6), and C(6,4). Then vectors \overrightarrow{AB}=\left\langle 2,4\right\rangle and \overrightarrow{AC}=\left\langle 4,2\right\rangle . The area of the triangle is half the magnitude of the cross product of these two vectors. \dfrac{1}{2}\left| \left| \begin{array}{l} {i\ \ j\ \ k}\\{2\ \ 4\ \ 0}\\{4\ \ 2\ \ 0} \end{array}\right|\right|=\dfrac{1}{2}\left| -12K\right|=\dfrac{1}{2}\cdot12=6.

Like in other solutions, we find that the three points of intersection are (2,2), (4,6), and (6,4). By the Pythagorean theorem, this is an isosceles triangle with base 2\sqrt{2} and equal length 2\sqrt{5}. The area of an isosceles triangle with base b and equal length l is \frac{b\sqrt{4l^2-b^2}}{4}. Plugging in b=2\sqrt{2} and l=2\sqrt{5}, \frac{2\sqrt{2}\cdot\sqrt{80-8}}{4}=\frac{\sqrt{576}}{4}=\frac{24}{4}=6.

Like in other solutions, we find the three points of intersection. Label these A(2,2), B(4,6), and C(6,4). By the Pythagorean Theorem, AB=AC=2\sqrt{5} and BC=2\sqrt{2}. By the Law of Cosines, \cos A=\frac{(2\sqrt{5})^2+(2\sqrt{5})^2-(2\sqrt{2})^2}{2\cdot2\sqrt{5}\cdot2\sqrt{5}}=\frac{20+20+8}{40}=\frac{32}{40}=\frac{4}{5} Therefore, \sin A=\sqrt{1-\cos^2 A}=\frac{3}{5}. By the extended Law of Sines 2R=\frac{a}{\sin A}=\frac{2\sqrt{2}}{\frac{3}{5}}=\frac{10\sqrt{2}}{3}R=\frac{5\sqrt{2}}{3},Then the area is \frac{abc}{4R}=\frac{2\sqrt{2}\cdot2\sqrt{5}^2}{4\cdot\frac{5\sqrt{2}}{3}}=6.

The area of a triangle formed by three lines, a_1x+a_2y+a_3=0,b_1x+b_2y+b_3=0, c_1x+c_2y+c_3=0 is the absolute value of \frac{1}{2}\cdot\frac{1}{(b_1c_2-b_2c_1)(a_1c_2-a_2c_1)(a_1b_2-a_2b_1)}\cdot \left| \begin{array}{l} {a_{1\ \ }a_{2}\ \ a_{3}}\\{b_{1\ \ }b_{2}\ \ b_{3}}\\{c_{1\ \ }c_{2}\ \ c_{3}} \end{array}\right|^2 Plugging in the three lines, -x+2y-2=0-2x+y+2=0x+y-10=0 the area is the absolute value of \frac{1}{2}\cdot\frac{1}{(-2-1)(-1-2)(-1+4)}\cdot \left| \begin{array}{l} {-1\ \ \ 2\ \ -2}\\{-2\ \ \ 1\ \ \ \ \ 2}\\{\ \ 1\ \ \ \ 1\ -10} \end{array}\right|^2=\frac{1}{2}\cdot\frac{1}{27}\cdot{18}^2=6

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