2019 AMC 10A Problem 7

Through $(2,2)$ the lines are $y = \tfrac12 x + 1$ and $y = 2x - 2$.

Meet each with $x + y = 10$, i.e. $y = 10 - x$:
- $\tfrac12 x + 1 = 10 - x$ gives $\tfrac32 x = 9$, so $x = 6$: the point $(6, 4)$.
- $2x - 2 = 10 - x$ gives $3x = 12$, so $x = 4$: the point $(4, 6)$.

The triangle has vertices $(2,2)$, $(6,4)$, $(4,6)$. The side on $x + y = 10$ runs from $(6,4)$ to $(4,6)$ and has length $\sqrt{2^2 + 2^2} = 2\sqrt2$. The distance from $(2,2)$ to the line $x + y - 10 = 0$ is
$$\frac{|2 + 2 - 10|}{\sqrt{1^2 + 1^2}} = \frac{6}{\sqrt2} = 3\sqrt2.$$
Area $= \tfrac12 \cdot 2\sqrt2 \cdot 3\sqrt2 = 6$.

The answer is $\boxed{\textbf{(C)}\ 6}$.

A triangle cut out by three lines is found by intersecting the lines pairwise. Once the vertices are known, pick the most convenient base: here one side lies on a line of slope $-1$, so its length and the height are both easy. The shoelace formula is the fallback when no side is convenient.

Shoelace on $(2,2), (6,4), (4,6)$:
$$\tfrac12\left|2\cdot4 + 6\cdot6 + 4\cdot2 - (2\cdot6 + 4\cdot4 + 6\cdot2)\right| = \tfrac12|8 + 36 + 8 - 12 - 16 - 12| = \tfrac12 \cdot 12 = 6.$$

• Arithmetic slips when solving for the intersections, or assuming the lines with slopes 1/2 and 2 are perpendicular (their slopes multiply to 1, not -1).
• Using the distance from $(2,2)$ to the line as $6$ (forgetting to divide by $\sqrt2$), which gives $6\sqrt2$, choice (E).