2019 AMC 12A Problem 5

Below is the professionally curated solution for Problem 5 of the 2019 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AMC 12A solutions, or check the answer key.

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Concepts:coordinate geometry triangle area shoelace formula

Difficulty rating: 1280

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 $x + y = 10?$

$4$

$4\sqrt{2}$

$6$

$8$

$6\sqrt{2}$

Solution:

The two lines are $y = \tfrac{1}{2}x + 1$ and $y = 2x - 2.$ Intersecting each with $x + y = 10$ gives the points $(6, 4)$ and $(4, 6).$

The triangle has vertices $(2, 2),$ $(6, 4),$ and $(4, 6).$ By the shoelace formula,

$\tfrac{1}{2}\left| 2(4 - 6) + 6(6 - 2) + 4(2 - 4) \right| = \tfrac{1}{2}\left| -4 + 24 - 8 \right| = 6.$

Thus, the correct answer is C.

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