Problem 99 - Intermediate Algebra

Question

Evaluate each expression for the given values. See Example 10. $$ \begin{aligned} &\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \text { for } x_{1}=-2, x_{2}=4\\\ &y_{1}=4, y_{2}=-4 \end{aligned} $$

Step-by-Step Solution

Answer

The value of the expression is 10.

Step 1: Identify the Formula

The expression provided is the distance formula, which is \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). This formula is used to calculate the distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) on a coordinate plane.

Step 2: Substitute the Given Values

Insert the provided values into the formula: \( x_1 = -2 \), \( x_2 = 4 \), \( y_1 = 4 \), and \( y_2 = -4 \). Substitute these into the equation: \[\sqrt{(4 - (-2))^2 + (-4 - 4)^2}\]

Step 3: Simplify Inside the Parentheses

Calculate the expressions inside the parentheses:\( x_2 - x_1 = 4 - (-2) = 4 + 2 = 6 \)\( y_2 - y_1 = -4 - 4 = -8 \)

Step 4: Square the Results

Square the results from the previous step:\( (x_2 - x_1)^2 = 6^2 = 36 \)\( (y_2 - y_1)^2 = (-8)^2 = 64 \)

Step 5: Add the Squared Values

Add the squared values together:\( 36 + 64 = 100 \)

Step 6: Take the Square Root

Take the square root of the sum calculated in the previous step to find the distance:\( \sqrt{100} = 10 \)

Key Concepts

Understanding the Coordinate Plane

The coordinate plane is a two-dimensional surface where we can plot points, lines, and curves. It's like a map that helps us understand where things are located on it.
You have two axes: the horizontal axis, which is the x-axis, and the vertical axis, which is the y-axis. These axes intersect at a point called the origin, labeled as \(0,0\).
Every point on the plane is identified by an ordered pair \( (x, y) \), meaning that x indicates how far along the horizontal axis the point is, and y shows how far along the vertical axis it is.

• The x-value tells us how far left or right to move from the origin.
• The y-value tells us how far up or down to move.

Understanding this setup is crucial for using algebra to find distances on the plane.

The Role of Algebra in Distance Calculation

Algebra is a branch of mathematics that deals with symbols and the rules for manipulating those symbols. In the context of a coordinate plane, algebra comes into play when we use formulas, like the distance formula, to find exact measurements between points.
In the distance formula, we use algebra to make calculations easier by working with the changes in x-values and y-values between two points. Here's how algebra simplifies this process:

• The expression \(x_2 - x_1\) finds the horizontal difference between two points.
• The expression \(y_2 - y_1\) finds the vertical difference.
• These differences are squared to ensure we only deal with positive numbers, as actual distance cannot be negative.
• The sum of these squares gives us the square of our distance, which we then root to find the actual distance.

By understanding algebra, you can easily handle these calculations, even with more complex numbers.

How to Calculate Distance on a Coordinate Plane

Calculating distance on a coordinate plane might seem tricky at first, but with the distance formula, it becomes straightforward. This formula helps you find how far apart two points are, using their coordinates.
Let's break it down step by step:

• Start by identifying your points, labeled as \( (x_1, y_1) \) and \( (x_2, y_2) \).
• Using the formula \[\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]\, substitute your point values in for the appropriate variables.
• Simplify the expressions inside the parentheses to find the horizontal and vertical changes.
• Square these results and add them—this represents the squared distance.
• Take the square root to find the actual distance between your points.

With practice, performing these steps becomes second nature, and you’ll be able to quickly determine distances in all sorts of problems.