We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

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