Problem 80 - Algebra A Combined Function

Question

Find the distance between each pair of points. Give an exact distance and a three-decimal-place approximation. (3,-2) and (-4,1)

Step-by-Step Solution

Answer

The exact distance is \( \sqrt{58} \) and approximately 7.616.

Step 1: Identify the Points

We are given two points: Point A at \((3, -2)\) and Point B at \((-4, 1)\).

Step 2: Use the Distance Formula

To find the distance between the two points, we use the distance formula:\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Substitute the given points into the formula, where \( (x_1, y_1) = (3, -2) \) and \( (x_2, y_2) = (-4, 1) \).

Step 3: Calculate the Differences

Calculate the differences for the formula:\( x_2 - x_1 = -4 - 3 = -7 \)\( y_2 - y_1 = 1 + 2 = 3 \).

Step 4: Calculate Squared Differences

Next, we calculate the squared differences:\((-7)^2 = 49\)\(3^2 = 9\).

Step 5: Add Squared Differences

Add the squared differences:\( 49 + 9 = 58 \).

Step 6: Find the Square Root

Take the square root of the sum to find the distance:\( d = \sqrt{58} \).

Step 7: Find a Decimal Approximation

To approximate, calculate \( \sqrt{58} \) using a calculator:\( \sqrt{58} \approx 7.616 \).

Key Concepts

Coordinate Geometry: Understanding the Distance Between Points

Coordinate geometry, often referred to as the geometry of graphs and coordinates, is a crucial part of understanding the mathematical layout of our world. By using points on a plane determined by their coordinate pairs, we can resolve various geometric problems, such as finding distances. In this specific problem, you use the distance formula to calculate the distance between two points.
This formula derives from the Pythagorean theorem and provides a systematic way to measure how far apart two points are in a two-dimensional space.
The general form of the distance formula is:

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

Here,

• \(x_1\) and \(y_1\) are the coordinates of the first point,
• \(x_2\) and \(y_2\) are the coordinates of the second point.

By plug-in these coordinates into the formula, you can calculate the straight-line distance, as demonstrated in the example with points (3, -2) and (-4, 1).

Square Root Calculation: Unlocking the Formula’s Result

As you navigate through the process of applying the distance formula, reaching the calculation of square roots becomes essential. Square root calculations stem from the need to find a number which, when multiplied by itself, results in a given value.
In this case, we had calculated the sum of squared differences, which was 58, and needed to determine its precise square root to finalize our distance measure.
This can be challenging when performed manually, as it frequently results in an irrational number that can’t be fully expressed as a simple fraction.

• The formula reaching \( \sqrt{58} \) exemplifies this challenge.
• It’s often practical to use calculators or computational tools to find such roots, especially when exact calculations are necessary for further applications.

Understanding square roots thus enables students to confidently transition from analytical expressions into comprehensible values, crucial for many areas of math.

Decimal Approximation: Bringing Precision to Everyday Calculations

When solving mathematical problems, you’ll often encounter numbers that can be cumbersome in their exact form. This is where decimal approximation steps in as an invaluable tool.
Specifically, when dealing with roots or irrational numbers, we can use approximations to arrive at a more manageable figure.

• For instance, calculating \( \sqrt{58} \) yields an irrational number which continues indefinitely.
• Using calculators, we approximate it to a more perceptible form like 7.616, enabling practical application or further processing.

Decimal approximation helps to simplify complex numbers and offers a balanced compromise between exactness and usability, ensuring results that are both accurate and easy to interpret. This approach is particularly useful in situations requiring precision but allows for practical limits.