Problem 14 - Prealgebra

Question

ZOO Beth is looking at a map of the zoo that is laid out on a coordinate system. Beth is at \((1,-1) .\) The gorilla house is at \((-2,-4)\) and the reptile exhibit is at \((3,2) .\) Is Beth closer to the gorilla house or the reptile exhibit?

Step-by-Step Solution

Answer

Beth is closer to the reptile exhibit.

Step 1: Identify the Points

Beth's position is at the coordinates \((1, -1)\), the gorilla house is at \((-2, -4)\), and the reptile exhibit is at \((3, 2)\). We will calculate how far Beth is from each location to determine which one is closer.

Step 2: Calculate Distance to Gorilla House

Use the distance formula to find the distance between Beth and the gorilla house. The formula is:\[\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]Substitute \((x_1, y_1) = (1, -1)\) and \((x_2, y_2) = (-2, -4)\):\[\sqrt{((-2) - 1)^2 + ((-4) - (-1))^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18}\]

Step 3: Simplify Distance to Gorilla House

Simplify \(\sqrt{18}\) to find the exact distance:\[\sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2}\]

Step 4: Calculate Distance to Reptile Exhibit

Use the distance formula again to find the distance between Beth and the reptile exhibit. Substitute \((x_1, y_1) = (1, -1)\) and \((x_2, y_2) = (3, 2)\):\[\sqrt{((3) - 1)^2 + ((2) - (-1))^2} = \sqrt{(2)^2 + (3)^2} = \sqrt{4 + 9} = \sqrt{13}\]

Step 5: Compare Distances

Now compare the distances: the distance to the gorilla house is \(3\sqrt{2}\), and the distance to the reptile exhibit is \(\sqrt{13}\). Calculate the approximate values:- For \(3\sqrt{2}\), approximate \(\sqrt{2} \approx 1.414\), so \(3 \times 1.414 \approx 4.242\).- For \(\sqrt{13}\), approximate \(\sqrt{13} \approx 3.606\).As \(3.606 < 4.242\), Beth is closer to the reptile exhibit.

Key Concepts

The Coordinate System

In the world of geometry and mapping, the coordinate system is an essential tool. It helps us pinpoint exact locations using a pair of numbers. These numbers, called coordinates, are represented as

• \((x, y)\)

The \(x\)-coordinate tells us how far left or right a point is from the origin, while the \(y\)-coordinate indicates how up or down the point is. You can think of the coordinate system as a big grid. The origin, where both coordinates are zero, is the point \((0, 0)\). This grid is incredibly useful for mapping and determining distances between points, just like in our example with Beth in the zoo.
When we say Beth is at \((1, -1)\), it means she's positioned 1 unit to the right and 1 unit down from the origin. Understanding this layout allows us to use various formulas to calculate distances precisely, without having to walk them out in real life.

Geometry and Distance

Geometry doesn't just help us understand shapes and angles; it's also crucial for calculating distances. When you're trying to find out how far apart two points are on a coordinate plane, you use the distance formula. It's derived from the Pythagorean theorem, which relates to the hypotenuse of a right triangle:

• \[ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Imagine a right triangle being formed by the line between Beth's position and one of the exhibits. The distances along the \(x\)-axis and \(y\)-axis make up the two legs of the triangle. By using this formula, you essentially calculate the hypotenuse, which is the direct distance between Beth and the point of interest.
In our example, we applied this formula twice: once to find how far Beth is from the gorilla house and again for the reptile exhibit. This approach gives us precise calculations that geometric intuition alone can’t provide.

Problem-Solving with Distances

Problem-solving in geometry often involves comparing distances to make decisions. The process is methodical, starting with identifying your points of interest on the coordinate system. After determining the coordinates of each point, it's important to apply the distance formula accurately.
When we calculated the distances to the gorilla house and the reptile exhibit, we gathered numerical results that needed further simplification and approximation to compare them clearly. This step is vital because exact results confirm our decision-making.
In this problem, approximating

• \( 3\sqrt{2} \)
• \( \sqrt{13} \)

helped reveal that Beth is closer to the reptile exhibit. This method of using math to arrive at conclusions is typical in problem-solving, offering a structured way to handle more complex questions that you might encounter as you advance in geometry.