2019 AMC 10A Problem 7

Below is the professionally curated solution for Problem 7 of the 2019 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AMC 10A solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:coordinate geometrytriangle area

Difficulty rating: 1280

7.

Two lines with slopes 12\frac{1}{2} and 22 intersect at (2,2).(2, 2). What is the area of the triangle enclosed by these two lines and the line x+y=10?x + y = 10?

44

424\sqrt{2}

66

88

626\sqrt{2}

Solution:

Let us first find the equations of the two lines. Using slope-intercept form, we know they are the form y=ax+b.y = ax + b.

For the first line, we know that a=12,a = \dfrac{1}{2}, so we get 2=122+b 2 = \dfrac{1}{2} \cdot 2 + b b=1. b = 1.

Similarly, for the second line, we get that a=2,a = 2, which gives us 2=22+b 2 = 2 \cdot 2 + b b=2. b = -2.

Our two lines are now y=12x+1y = \dfrac{1}{2}x + 1 and y=2x2.y = 2x - 2.

We can rewrite x+y=10x + y = 10 as y=10x.y = 10 - x. Substituting this into the first line yields 10x=12x+1 10 - x = \dfrac{1}{2}x + 1 x=6,y=4. x = 6, y = 4.

Similarly, for the second line, 10x=2x2 10 - x = 2x - 2 x=4,y=6. x = 4, y = 6.

The three vertices of the triangle are therefore (2,2),(6,4),(2, 2), (6, 4), and (4,6).(4, 6).

Note that these vertices form an isosceles triangle (distance formula yields the three sides as 25,25,2\sqrt{5}, 2\sqrt{5}, and 22.2\sqrt{2}.

The midpoint of the base is (5,5),(5, 5), and applying the distance formula again tells us that the height is 32.3\sqrt{2}.

The area is therefore 122232=6. \dfrac{1}{2} \cdot 2\sqrt{2} \cdot 3 \sqrt{2} = 6. Thus, C is the correct answer.

Problem 7 in Other Years