Problem 16 - College Algebra with Modeling and Visualization

Question

Find the exact distance between the two points. Where appropriate, also give approximate results to the nearest hundredth. $$ (-1,-6),(-8,-5) $$

Step-by-Step Solution

Answer

The exact distance is \(5\sqrt{2}\), approximately 7.07.

Step 1: Identify the Points

The points provided are \( x_1 = -1, y_1 = -6 \) and \( x_2 = -8, y_2 = -5 \). These are the coordinates of the two points between which we need to find the distance.

Step 2: Recall the Distance Formula

The distance \(d\) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula: \\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Step 3: Substitute the Values in the Formula

Plug the coordinates into the distance formula:\[ d = \sqrt{(-8 - (-1))^2 + (-5 - (-6))^2} \] This simplifies to:\[ d = \sqrt{(-8 + 1)^2 + (-5 + 6)^2} \]

Step 4: Simplify Inside the Square Root

Calculate the values inside the parentheses:\[ d = \sqrt{(-7)^2 + 1^2} \] Further simplifying, we have:\[ d = \sqrt{49 + 1} \]

Step 5: Simplify and Calculate the Square Root

Add the values under the square root:\[ d = \sqrt{50} \] which simplifies to the exact form:\[ d = 5\sqrt{2} \]

Step 6: Find the Approximate value

To find the approximate distance to the nearest hundredth, calculate the decimal value of \(5\sqrt{2}\). This gives:\[ d \approx 7.07 \]

Key Concepts

Coordinate Geometry

Coordinate geometry, also known as analytical geometry, is a branch of mathematics that deals with defining and using coordinates to specify the positions of points and other geometric entities. In coordinate geometry, we usually use the Cartesian coordinate system which consists of two axes: the x-axis (horizontal) and the y-axis (vertical).

In this system, each point is defined by a pair of numerical coordinates, written as \((x, y)\). The first number \(x\) represents the horizontal position, while the second number \(y\) indicates the vertical position. This allows us to visualize and calculate various geometric properties and relationships.

The distance formula is a key concept in coordinate geometry. It helps in finding the distance between two points on a Cartesian plane. The formula is derived from the Pythagorean theorem and is crucial for solving problems related to distances in a two-dimensional space.

Square Root Simplification

Square root simplification is the process of breaking down a square root into its simplest form. When evaluating expressions, especially those involving the distance formula, it's important to simplify the square roots to make calculations easier.

For instance, in the exercise, you find \(d = \sqrt{50}\) as an intermediate step in the distance calculation. Simplifying this involves recognizing that 50 can be expressed as 25 multiplied by 2, which are perfect squares. Therefore, \(\sqrt{50}\) can be rewritten as \(\sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2}\).

• The square root of 25 is 5, hence \(\sqrt{25} = 5\).
• This leads to \(d = 5\sqrt{2}\), which is the exact form.

Simplifying square roots not only obeys mathematical clarity but also assists in further approximations if needed.

Approximate Calculation

Approximate calculation is used when you need a numerical value that doesn't need to be exact but is sufficiently close for practical purposes. After determining the exact distance between two points, you might find it useful to convert the exact expression into a decimal approximation.

In this exercise, after arriving at the exact distance of \(d = 5\sqrt{2}\), the next step involves calculating its approximate value. To do this, you can estimate \(\sqrt{2}\) as approximately 1.414, and multiply it by 5, resulting in:

\[d \approx 5 \times 1.414 = 7.07\]

Using approximation is beneficial, especially when you want a more intuitive understanding of the magnitude or need to communicate results in a context where precision is flexible. Keep in mind that each context may require a different level of approximation.