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.

Logic Dec 8, 2018

#1**+2 **

I think the best thing to do here is to construct a diagram. The problem gives you vertices with coordinates from the Cartesian plane, and it even gives you an inequality. Here is the construction I created. I will reference it throughout the solving process:

I first located the given coordinates and connected them with vertical and horizontal lines. I then graphed the inequality\(x>7y\), which can be rewritten as \(y<\frac{x}{7}\) . I added a few points, as they will become relevant when I solve.

One way to solve this problem is to figure out the ratio of the area created by the inequality \(y<\frac{x}{7}\) and the sides of the rectangle to the area of the entire rectangle. I will do this by first finding the coordinates of the point E. Point E is located on \(\overline{BC}\) , so its x-coordinate is also 2009. If you know the x-coordinate, then the y-coordinate of any point on \(y=\frac{x}{7}\) is \(\frac{x}{7}\) . Therefore, the y-coordinate is \(\frac{2009}{7}\) .

The formula for the area of the triangle is \(\frac{1}{2}bh\) . \(\triangle ABE\) is a right triangle, so the base and height are the side lengths that are not the hypotenuse. We know the length of the base, \(AB\) , because it is in the diagram. The height is the length of \(BE\) , or \(\frac{2009}{7}\)

\(A_{\triangle}=\frac{1}{2}*AB*BE=\frac{1}{2}*2009*\frac{2009}{7}=\frac{2009^2}{14}\)

We can also calculate the area of the rectangle

\(A_{\text{rect.}}=bh=AB*BC=2009*2010\)

As aforementioned, the probability of (x,y) such that x>7y is just the ratio of these areas.

\(P(x,y|x>7y)=\frac{A_{\triangle}}{A_{\text{rect.}}}\) | As aforementioned, the probability of (x,y) such that x>7y is just the ratio of these areas. Substitute in the values and simplify. |

\(P(x,y|x>7y)=\frac{\frac{2009^2}{14}}{2009*2010}\) | Use algebraic manipulation to simplify this. |

\(P(x,y|x>7y)=\frac{2009^2}{14*2009*2010}\) | |

\(P(x,y|x>7y)=\frac{2009}{14*2010}\) | At this point, you must use a calculator. |

\(\)\(P(x,y|x>7y)=0.07139\) | |

TheXSquaredFactor Dec 8, 2018

#1**+2 **

Best Answer

I think the best thing to do here is to construct a diagram. The problem gives you vertices with coordinates from the Cartesian plane, and it even gives you an inequality. Here is the construction I created. I will reference it throughout the solving process:

I first located the given coordinates and connected them with vertical and horizontal lines. I then graphed the inequality\(x>7y\), which can be rewritten as \(y<\frac{x}{7}\) . I added a few points, as they will become relevant when I solve.

One way to solve this problem is to figure out the ratio of the area created by the inequality \(y<\frac{x}{7}\) and the sides of the rectangle to the area of the entire rectangle. I will do this by first finding the coordinates of the point E. Point E is located on \(\overline{BC}\) , so its x-coordinate is also 2009. If you know the x-coordinate, then the y-coordinate of any point on \(y=\frac{x}{7}\) is \(\frac{x}{7}\) . Therefore, the y-coordinate is \(\frac{2009}{7}\) .

The formula for the area of the triangle is \(\frac{1}{2}bh\) . \(\triangle ABE\) is a right triangle, so the base and height are the side lengths that are not the hypotenuse. We know the length of the base, \(AB\) , because it is in the diagram. The height is the length of \(BE\) , or \(\frac{2009}{7}\)

\(A_{\triangle}=\frac{1}{2}*AB*BE=\frac{1}{2}*2009*\frac{2009}{7}=\frac{2009^2}{14}\)

We can also calculate the area of the rectangle

\(A_{\text{rect.}}=bh=AB*BC=2009*2010\)

As aforementioned, the probability of (x,y) such that x>7y is just the ratio of these areas.

\(P(x,y|x>7y)=\frac{A_{\triangle}}{A_{\text{rect.}}}\) | As aforementioned, the probability of (x,y) such that x>7y is just the ratio of these areas. Substitute in the values and simplify. |

\(P(x,y|x>7y)=\frac{\frac{2009^2}{14}}{2009*2010}\) | Use algebraic manipulation to simplify this. |

\(P(x,y|x>7y)=\frac{2009^2}{14*2009*2010}\) | |

\(P(x,y|x>7y)=\frac{2009}{14*2010}\) | At this point, you must use a calculator. |

\(\)\(P(x,y|x>7y)=0.07139\) | |

TheXSquaredFactor Dec 8, 2018