Problem 13 - College Algebra

Question

Find the distance between each pair of points. If necessary, round answers to two decimals places. $$(0,-\sqrt{3})\( and \)(\sqrt{5}, 0)$$

Step-by-Step Solution

Answer

The distance between the two points is approximately 2.83.

Step 1: Identify the coordinates

First identify the coordinates of the two points. We have point A at coordinates (0, -\(\sqrt{3}\)) and point B at coordinates (\(\sqrt{5}\), 0).

Step 2: Substitute into Distance Formula

Next, substitute these values into the distance formula: \[d = \sqrt{(\sqrt{5} - 0)^2 + (0 - -\sqrt{3})^2}\]

Step 3: Simplify the expression inside the square root

We get: \[d = \sqrt{5 + 3}\]

Step 4: Simplify the Square Root

\[d = \sqrt{8}\] Now, the square root of 8 can be simplified as \(2\sqrt{2}\) because 8 can be written as 4*2, and the square root of 4 is 2.

Step 5: Round the answer

The question requires that the answer be rounded to two decimal places if necessary. The decimal form of \(2\sqrt{2}\) is approximately 2.83. So, the distance between the two points to two decimal places is 2.83.

Key Concepts

Coordinate Geometry

Coordinate geometry is a branch of geometry where points are defined in a coordinate plane. It allows you to use algebra to solve geometric problems. In our exercise, we're asked to find the distance between two points, specifically

• (0, -\(\sqrt{3}\))
• (\(\sqrt{5}\), 0)

To find this distance, we use the distance formula, which connects the algebraic representation of points with geometric concepts.The distance formula \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\] gives the straight-line distance between any two points

• \((x_1, y_1)\)
• \((x_2, y_2)\)

This formula comes from the Pythagorean theorem applied in the coordinate plane.This makes coordinate geometry a powerful tool for calculations and proofs.

Square Roots

The concept of square roots is fundamental in finding distances in a coordinate plane. A square root of a number is a value that, when multiplied by itself, gives the original number. For example,

• the square root of 16 is 4

Because 4 times 4 is 16.In the exercise, square roots appear in the calculations. For instance, \[-\sqrt{3}\] and \[\sqrt{5}\] are given as part of the points' coordinates. When solving the distance formula, you'll often end up with square roots. They help you find exact distances, even when points have irrational number coordinates. Understanding how to work with square roots ensures precise and orderly calculations, essential when dealing with geometric formulas.

Simplifying Radicals

Simplifying radicals is the process of expressing square roots in their simplest form. This is crucial when working with distance calculations, as it allows for a cleaner and more understandable result. For the exercise, the expression \(\sqrt{8}\) was simplified to \(2\sqrt{2}\). Here's how that's done:

• Identify perfect square factors of 8, which is 4.
• The square root of 4 is 2.
• So, \(\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \cdot \sqrt{2} = 2\sqrt{2}\).

This technique helps streamline calculations and provides a more accurate result when further conversion or rounding is needed. Knowing how to simplify a radical is a vital skill in mathematics, helping to prevent errors in larger, more complex computations.

Rounding Decimals

Rounding decimals is often necessary to provide a concise and practical result. After simplifying radicals, the next step involves converting that value into a decimal, which in this case is approximately \(2.83\). This rounding is crucial when a problem asks for a result to be presented in a particular format, such as two decimal places. Here is how rounding works:

• Identify the digit beyond your rounding point (third decimal place in this case).
• If this digit is 5 or greater, increase the second decimal place digit by one.

Rounding helps in providing answers that are easier to read and more applicable in real-world scenarios. Having a rounded distance to 2.83 makes it neater and more practical, especially in real-life applications like engineering and navigation, where precise yet manageable figures are needed.