### Instruction: Point $(x,y)$ is randomly picked from the rectangular region with vertices at $(0,0),(2009,0),(2009,2010),$ and $(0,2010)$. What is the probability that $x > 7y$? Express your answer as a common fraction.### Response: I want to find the area of the region where $x > 7y$ and divide it by the total area of the rectangle.
To do that, I need to find the points where the line $x = 7y$ intersects the rectangle.
If $y = 0$, then $x = 0$, so one point is $(0,0)$.
If $x = 2009$, then $y = \frac{2009}{7}$, so another point is $(2009,\frac{2009}{7})$.
If $y = 2010$, then $x = 7 \times 2010 = 14070$, which is outside the rectangle, so there is no intersection on the top edge.
If $x = 0$, then $y = 0$, which we already have, so there is no other intersection on the left edge.
So the region where $x > 7y$ is a triangle with vertices at $(0,0),(2009,0),$ and $(2009,\frac{2009}{7})$.
The area of this triangle is $\frac{1}{2} \times 2009 \times \frac{2009}{7} = \frac{2009^2}{14}$.
The area of the rectangle is $2009 \times 2010$.
So the probability that $x > 7y$ is $\frac{\frac{2009^2}{14}}{2009 \times 2010} = \frac{2009}{14 \times 2010}$.