Problem 86 - Intermediate Algebra

Question

Find the distance between each pair of points. Give an exact distance and a three-decimal-place approximation. See Example 6 $$ (9.6,2.5) \text { and }(-1.9,-3.7) $$

Step-by-Step Solution

Answer

The exact distance is \(\sqrt{170.69}\), approximately 13.065.

Step 1: Identify the coordinates

First, identify the coordinates of the given points. The first point is \((9.6, 2.5)\) and the second point is \((-1.9, -3.7)\).

Step 2: Use the distance formula

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

Step 3: Substitute the coordinates into the formula

Substitute the coordinates \((9.6, 2.5)\) and \((-1.9, -3.7)\) into the distance formula:\[d = \sqrt{((-1.9) - 9.6)^2 + ((-3.7) - 2.5)^2}\]

Step 4: Calculate the differences

Calculate \((-1.9) - 9.6\) and \((-3.7) - 2.5\):\[(-1.9) - 9.6 = -11.5\]\[(-3.7) - 2.5 = -6.2\]

Step 5: Square the differences

Square the differences calculated:\[(-11.5)^2 = 132.25\]\[(-6.2)^2 = 38.44\]

Step 6: Sum the squares

Add the squares of the differences:\[132.25 + 38.44 = 170.69\]

Step 7: Calculate the square root

Calculate the square root of the sum to find the exact distance:\[d = \sqrt{170.69} = 13.06540162\]

Step 8: Round the distance

Round the exact distance to three decimal places: 13.065.

Key Concepts

Points and Coordinates

To calculate the distance between two points on a plane, it's essential first to understand what points and coordinates are. Points are specific locations on a coordinate plane, represented by pairs of numbers, often termed as coordinates. In a coordinate system, these points are denoted as \(x, y\), where \(x\) signifies the horizontal position and \(y\) indicates the vertical position.
This arrangement allows us to map any location in a two-dimensional space. For instance, in our problem, we have two points: \( (9.6, 2.5)\), where 9.6 is the x-coordinate, and 2.5 is the y-coordinate, and \( (-1.9, -3.7)\), where -1.9 is the x-coordinate, and -3.7 is the y-coordinate.
Recognizing these coordinates is the first step in solving any distance-related problem using the distance formula. Thus, knowing how to accurately identify and place these points on a graph or plane is critical in various algebraic methods, including calculating distances.

Square Root Calculation

The concept of square root plays a significant role in determining the distance between two points. When we employ the distance formula \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\], the expression under the square root, known as the radicand, must be simplified to solve for \(d\).
During calculations, first, you square the differences of x-coordinates and y-coordinates. \[ (x_2 - x_1)^2 \] and \[ (y_2 - y_1)^2 \] are calculated individually.

• In our example, \((-11.5)\) and \((-6.2)\) are squared to get 132.25 and 38.44 respectively

This means we need to find the square root of 170.69. Calculating the square root is a method of finding a number which, when multiplied by itself, gives a specified number. This is essential for achieving the exact distance between two points.
Square root calculations can be done using calculators, as they might involve irrational numbers, making mental computation less feasible. Remember, precision in square root calculations is crucial, especially when you're asked for distance to a specific decimal point.

Algebra Problem-Solving

Solving algebra problems involves juggling with equations and formulas, similar to our distance problem. Often, you'll be required to substitute known values into formulas and solve for unknowns.
Here's how we tackled this using algebra:

• Firstly, identified known coordinates and substitute them into the distance formula.
• This involves solving two primary algebraic calculations: differences between coordinates, squaring them, and summing those squares.
• Lastly, by finding the square root of their sum, we determine the precise distance.

This problem-solving approach is algebraic because it involves applying arithmetic and algebraic operations sequentially to reach the solution. It demands understanding the operations, manipulating numbers from one form to another, and consistently checking if each step aligns with algebraic principles.
When solving algebra problems, it’s also important to work systematically—breaking down larger problems into more manageable steps ensures accuracy and clarity. Algebraic problem-solving in this context prepares you for tackling not only geometrical problems but also real-world mathematical challenges.