Problem 83 - Intermediate Algebra

Question

Find the distance between each pair of points. Give an exact distance and a three-decimal-place approximation. See Example 6 $$ (0,-\sqrt{2}) \text { and }(\sqrt{3}, 0) $$

Step-by-Step Solution

Answer

The exact distance is \(\sqrt{5}\), and the decimal approximation is 2.236.

Step 1: Understand the Problem

We need to find the distance between two points: \((0, -\sqrt{2})\) and \((\sqrt{3}, 0)\). The task requires us to provide both an exact distance and a decimal approximation.

Step 2: Use the Distance Formula

The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) in a plane is given by the distance formula: \[D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\] Substitute the given points into the formula.

Step 3: Substitute the Values into the Formula

Substitute \(x_1 = 0\), \(y_1 = -\sqrt{2}\), \(x_2 = \sqrt{3}\), and \(y_2 = 0\) into the distance formula:\[D = \sqrt{(\sqrt{3} - 0)^2 + (0 - (-\sqrt{2}))^2}\]This simplifies to: \[D = \sqrt{(\sqrt{3})^2 + (\sqrt{2})^2}\]

Step 4: Simplify the Equation

Calculate the squares of the square roots: \[(\sqrt{3})^2 = 3 \quad \text{and} \quad (\sqrt{2})^2 = 2\] Substitute back into the equation: \[D = \sqrt{3 + 2}\]Simplify further:\[D = \sqrt{5}\]

Step 5: Approximate the Distance

Using a calculator, find the decimal approximation of \(\sqrt{5}\): \[\sqrt{5} \approx 2.236\] Round the approximation to three decimal places as required.

Key Concepts

Distance Between Points

In geometry, calculating the distance between two points on a plane is a common task. To achieve this, we use the distance formula, which is derived from the Pythagorean theorem.
This formula can be written as:

• \( D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

Here, \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points. By substituting the coordinate values into the formula, we can determine the linear distance that spans directly between them.
This formula efficiently calculates the hypotenuse of a right triangle formed by the coordinate axes, helping us measure distances on a two-dimensional plane.

Exact Distance

When a problem requires an exact distance, we need to provide the result in its simplest form, usually as a mathematical expression. For the problem you’re facing, the exact distance is determined by substituting the points into the formula:\[D = \sqrt{(\sqrt{3} - 0)^2 + (0 - (-\sqrt{2}))^2}\]. This equation simplifies to \( \sqrt{5} \), providing the exact distance as an expression, without estimating it to a decimal format.
Using the exact distance is crucial in mathematical proofs or when solutions must retain maximum accuracy and precision.
It’s useful to represent exact distances in their radical forms to avoid the pitfalls of decimal approximations, which could potentially lead to inaccuracies in more complex calculations.

Decimal Approximation

Often, obtaining a decimal approximation of an exact distance is necessary for real-world applications. This means converting the radical form of the distance into a more tangible number.
Using a calculator, the square root of 5, which is our exact distance, approximates to 2.236. This is obtained by performing the square root operation on the calculator or using computational software.
Rounding to three decimal places aligns with standard practices for precision in scientific and engineering tasks.

• This helps in making the measurement more practical and easy to use in real-life scenarios.
• Also, it is essential when presenting data clearly and concisely.

Being adept at moving from exact expressions to decimal approximations is a key skill in both academics and various professional fields.

Squaring Square Roots

The process of squaring square roots within the distance formula can sometimes perplex students. When you square a square root, you essentially cancel out the square root operation.
This means:

• \((\sqrt{3})^2 = 3\)
• \((\sqrt{2})^2 = 2\)

This is because the squaring operation and the square root operation are inverse functions. When applying this to the distance formula, it simplifies the equation significantly, leading to more straightforward computation.
Understanding this inversion can make radical simplifications intuitive and stress-free in mathematical calculations. Emphasizing these basics helps build a solid foundation, on which more complex algebraic concepts can be layered.