2019 AMC 10A Problem 7

A triangle made by three lines has its corners where the lines cross, two at a time. So first find the three corners.

1. Write the two lines. A line through $(2,2)$ with slope $\tfrac12$ rises $1$ for every $2$ across. Going back $2$ from $(2,2)$ reaches $(0,1)$, so it is $y = \tfrac12 x + 1$. The line with slope $2$ through $(2,2)$ is $y = 2x - 2$ (check: $2 \cdot 2 - 2 = 2$).
2. Rewrite the third line. $x + y = 10$ is the same as $y = 10 - x$.
3. First corner. Set $\tfrac12 x + 1 = 10 - x$. Then $\tfrac32 x = 9$, so $x = 6$ and $y = 4$. Corner $(6,4)$.
4. Second corner. Set $2x - 2 = 10 - x$. Then $3x = 12$, so $x = 4$ and $y = 6$. Corner $(4,6)$.
5. Third corner. The two lines meet at $(2,2)$.

Now find the area of the triangle with corners $(2,2)$, $(6,4)$ and $(4,6)$.

6. Draw a box around it. The corners reach from $x = 2$ to $x = 6$ and from $y = 2$ to $y = 6$, so they fit in a $4 \times 4$ square with area $16$.
7. Cut off the three corner triangles of the box that are outside our triangle:
- between $(2,2)$ and $(6,4)$, below the triangle: a right triangle with legs $4$ and $2$, area $4$;
- between $(2,2)$ and $(4,6)$, left of the triangle: legs $2$ and $4$, area $4$;
- between $(6,4)$ and $(4,6)$, in the top-right corner of the box: legs $2$ and $2$, area $2$.
8. Subtract. $16 - 4 - 4 - 2 = 6$.

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

Three lines make a triangle whose corners are the pairwise crossing points, so the first job is always to solve three small pairs of equations. Once the corners are known, "box it and subtract the corners" finds the area of any triangle on a grid without needing a height.

Use the side on $x + y = 10$ as the base. It runs from $(6,4)$ to $(4,6)$: $2$ left and $2$ up, so its length is $\sqrt{2^2 + 2^2} = 2\sqrt2$. The height is the distance from $(2,2)$ to that line. The line $x + y = 10$ slants down at $45^\circ$, so the straight path from $(2,2)$ that meets it at a right angle goes up at $45^\circ$, along $(3,3), (4,4), (5,5)$. It reaches the line at $(5,5)$, a distance of $\sqrt{3^2 + 3^2} = 3\sqrt2$. So the area is $\tfrac12 \cdot 2\sqrt2 \cdot 3\sqrt2 = 6$.

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 $\boxed{\textbf{(C)}\ 6}$.