Problem 25 - Algebra 1

Question

Find the distance between the two points. Round the result to the nearest hundredth if necessary. $$\left(\frac{1}{3}, \frac{1}{6}\right),\left(-\frac{2}{3}, \frac{8}{3}\right)$$

Step-by-Step Solution

Answer

The distance between the points \((\frac{1}{3}, \frac{1}{6})\) and \(-\frac{2}{3}, \frac{8}{3}\) is approximately \(3.35\).

Step 1: Identify the Coordinates of the Two Points

The two points are \((\frac{1}{3}, \frac{1}{6})\) and \(-\frac{2}{3}, \frac{8}{3}\). So, we have x1 = \(\frac{1}{3}\), y1 = \(\frac{1}{6}\), x2 = \(-\frac{2}{3}\), and y2 = \(\frac{8}{3}\).

Step 2: Substituting the coordinates into the Distance Formula

Plugging these coordinates into the formula, we get: \[d = \sqrt{[(-\frac{2}{3} - \frac{1}{3})^2 + (\frac{8}{3} - \frac{1}{6})^2]}\]

Step 3: Simplifying Each Part of the Equation

First, simplify the terms inside the parentheses: \[d = \sqrt{[-\frac{3}{3}]^2 + [\frac{24}{6} - \frac{1}{6}]^2} = \sqrt{[(-1)^2 + (\frac{23}{6})^2]}\] Now, simplify the expression inside the square root: \[d = \sqrt{[1 + (\frac{115}{36})^2]} = \sqrt{1 + \frac{13225}{1296}} = \sqrt{\frac{1297}{1296} + \frac{13225}{1296}}\]

Step 4: Adding and Simplifying

Adding the terms together and taking the square root, we get the final answer: \[d = \sqrt{\frac{14522}{1296}} = \sqrt{11.20}\]

Key Concepts

Understanding Coordinates

Coordinates are essential in the world of geometry. They represent the position of a point in a plane using a pair of numbers. Understanding them is the first step to solving problems involving the distance formula. Let's break down how they work:

• The first number in the pair is called the x-coordinate. It tells us how far along the horizontal axis (left or right) the point is.
• The second number is the y-coordinate. This indicates how far along the vertical axis (up or down) the point is.

An example would be the point \( \left(\frac{1}{3}, \frac{1}{6}\right) \). Here, \(\frac{1}{3}\) is the x-coordinate and \(\frac{1}{6}\) is the y-coordinate. Similarly, for the point \(\left(-\frac{2}{3}, \frac{8}{3}\right)\), the x-coordinate is \(-\frac{2}{3}\) and the y-coordinate is \(\frac{8}{3}\).These coordinates allow us to accurately locate points on a two-dimensional grid, which is a fundamental concept when using the distance formula.

Simplification in Calculations

Simplification is a crucial step in solving mathematical problems, especially when working with fractions. It involves reducing fractions and expressions to make calculations more manageable. Here's what you need to keep in mind about simplification:

• When subtracting fractions, find a common denominator to combine them correctly. For example, \(\frac{8}{3} - \frac{1}{6} \) involves changing the fractions to have a common denominator of 6, resulting in \(\frac{24}{6} - \frac{1}{6} = \frac{23}{6}\).
• Simplify squared terms to their simplest form. For instance, squaring \(-1\) results in \(1\).
• Combine all simplified parts to achieve a cleaner expression, which makes taking square roots or further operations more straightforward.

In our example, each step of simplification helps make the computation under the square root sign clearer, leading us closer to finding the precise distance between the two points.

Calculating Distance Between Points

The distance formula is a fundamental tool in geometry for finding how far apart two points are. Here's how it works:The formula itself is:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]

• Subtract the x-coordinates of the two points to find the horizontal distance: \(-\frac{2}{3} - \frac{1}{3} = -1\)
• Subtract the y-coordinates to determine the vertical distance: \(\frac{8}{3} - \frac{1}{6} = \frac{23}{6}\).
• Square both these distances to ensure they are non-negative.
• Add the results and take the square root to find the distance \(d\).

In the example, these steps yield the distance:\[d = \sqrt{1 + \left(\frac{23}{6}\right)^2} = \sqrt{1 + \frac{529}{36}} = \sqrt{\frac{14522}{1296}}\]After a bit of extra calculation, this simplifies down to \(\sqrt{11.20}\), giving us a precise measure of the distance. Rounding the final result to the nearest hundredth, if necessary, gives the solution in a more usable form. Understanding this process helps in accurately solving these types of distance problems.