Problem 10 - Precalculus Essentials

Question

Find the distance between each pair of points. If necessary, express answers in simplified radical form and then round to two decimals places. $$(0,-2) \text { and }(4,3)$$

Step-by-Step Solution

Answer

The distance between the points (0,-2) and (4,3) is \(\sqrt{{41}}\), which is approximately 6.40 when rounded to two decimal places.

Step 1: Identify the Points

Take the two given points. The first point is (0,-2) and the second point is (4,3). Thus, \(x_1 =0\), \(y_1 = -2\), \(x_2 =4\) and \(y_2 =3\) are the values that will be substituted into the distance formula.

Step 2: Plug the Points into the Formula

Substitute the identified points into the distance formula, \(\sqrt{{(x_2 - x_1)^2 + (y_2 - y_1)^2}}\). This results in \(\sqrt{{(4 - 0)^2 + (3 - -2)^2}}\).

Step 3: Perform the Operations

Perform the subtraction and squaring operations in the formula. This yields \(\sqrt{{4^2 + 5^2}}\). Squaring each term, we get \(\sqrt{{16 + 25}}\).

Step 4: Simplify the Expression

Add the numbers together under the square root to simplify it, which results in \(\sqrt{{41}}\).

Step 5: Approximate the Result

Although the simplified radical form is exact, you can approximate to two decimal places by calculating the square root of 41, which results in approximately 6.40.

Key Concepts

Coordinate Geometry

Coordinate geometry involves plotting and analyzing points on a plane using their coordinates. A coordinate plane is divided into four quadrants by the x-axis and y-axis. Each point on this plane is represented by an ordered pair \(x, y\).

In our exercise, the points \((0,-2)\) and \((4,3)\) are located in different quadrants. By identifying the x and y values from each point, we can then use these to calculate distances or midpoints.

• The first point \((0,-2)\) has x-coordinate 0 and y-coordinate -2.
• The second point \((4,3)\) has x-coordinate 4 and y-coordinate 3.

Understanding how to work with these coordinates is crucial for solving problems related to distances, angles, and other spatial relationships on the plane.

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry that relates the sides of a right triangle. It's commonly expressed as \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse.

When applied to coordinate geometry, it becomes the distance formula, \(\sqrt{{(x_2 - x_1)^2 + (y_2 - y_1)^2}}\). This formula allows us to find the distance between two points on a plane.

In the exercise, after determining \((x_1, y_1)\) and \((x_2, y_2)\), we plug these values into the formula:

• \( (4-0)^2 = 16\)
• \((3-(-2))^2 = 25\)

By combining the results, we calculate \(\sqrt{41}\). This direct use of the Pythagorean Theorem is what simplifies complex geometrical relationships into manageable calculations.

Simplifying Radicals

Simplifying radicals is a process to make expressions involving square roots easier to handle. When given a square root like \(\sqrt{41}\), it is already in its simplest form if no perfect square factors exist within the radicand (the number under the square root).

To simplify radicals, you should:

• Factor the number inside the square root into its smallest prime numbers.
• Look for pairs of prime factors to bring outside of the square root.

With \(\sqrt{41}\), since 41 is a prime number, it cannot be simplified further, ensuring that \(\sqrt{41}\) is as simple as it gets.

This skill is useful in reducing complexities in mathematical expressions, leading to cleaner, more understandable solutions.