Problem 24 - Algebra 1

Question

Find the distance between the two points. Round the result to the nearest hundredth if necessary. $$\left(\frac{1}{2}, \frac{1}{4}\right),(2,1)$$

Step-by-Step Solution

Answer

The distance between the two points is approximately 1.68

Step 1: Identifying the coordinates

Identify the coordinates of the two points. First point is \((\frac{1}{2}, \frac{1}{4})\) so \(x_1 = \frac{1}{2}\) and \(y_1 = \frac{1}{4}\). Second point is \((2, 1)\) so \(x_2 = 2\) and \(y_2 = 1\).

Step 2: Substitution into the distance formula

Substitute these values into the distance formula: \(d = \sqrt{(2 - \frac{1}{2})^2 + (1 - \frac{1}{4})^2}\).

Step 3: Simplifying inside the square root

Simplify inside the square root to ease the calculation: \(d = \sqrt{(1.5)^2 + (0.75)^2}\).

Step 4: Calculation of the distance

Calculate the squares, sum them up and take the square root: \(d = \sqrt{(2.25) + (0.5625)} = \sqrt{2.8125}\).

Step 5: Simplifying the Result

Now, take the square root of \(2.8125\) to find the distance. The resulting distance is approximately \(1.68\), rounded to the nearest hundredth.

Key Concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is a fascinating branch of mathematics that simplifies geometry using a coordinate system. This enables us to employ algebraic methods to solve geometric problems. In this exercise, the coordinate plane consists of two perpendicular axes, the x-axis, and the y-axis.
Each point in the plane is defined by a pair of numerical coordinates, such as \(\left( \frac{1}{2}, \frac{1}{4}\right)\) and \( (2,1) \). Here, the first number in each pair is the x-coordinate, showing how far along the x-axis a point is. The second number is the y-coordinate, indicating a point's position along the y-axis.

• First Point: \(\left( \frac{1}{2}, \frac{1}{4}\right)\) where the x-coordinate is \(x_1 = \frac{1}{2}\) and the y-coordinate is \(y_1 = \frac{1}{4}\).
• Second Point: \( (2,1) \) where the x-coordinate is \(x_2 = 2\) and the y-coordinate is \(y_2 = 1\).

By utilizing these coordinates, we can explore various geometric concepts such as distance, slope, and midpoints. The exercise requires calculating the distance between these two identified points.

Algebra

Algebra functions as the engine that drives mathematical problem-solving, operating as the language through which we express and manipulate numbers and operations. In this exercise, algebra is the indispensable tool that facilitates finding the distance between two points using the distance formula.
The distance formula arises from the Pythagorean theorem and is expressed as: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]. This formula calculates the direct line (or hypotenuse) between two points in a two-dimensional space.

• Note the step-by-step process:
• Substitute each of the coordinates derived from coordinate geometry into the formula.
• The difference between x-coordinates and y-coordinates is first calculated separately.
• Each difference is then squared to eliminate any negative values.

This formula showcases the power of algebra as it transforms abstract numbers into actionable results, effortlessly calculating distances and facilitating geometric explorations.

Square Roots

The square root plays a crucial role in calculations involving geometry and algebra, as seen in this exercise. Understanding square roots helps in understanding the precision required by certain mathematical calculations.
The square root of a number is a value that, when multiplied by itself, gives the original number. It is depicted using the radical symbol \(\sqrt{}\). In the context of the distance formula, square roots are used to "unsquare" the results of the squared differences.

• After calculating \( (x_2 - x_1)^2 \) and \( (y_2 - y_1)^2 \), the sum of these values is subjected to the square root operation.
• The exercise narrows down to the square root of \(2.8125\), simplifying the distance calculation.
• This results in approximately \(1.68\), which is then rounded to the nearest hundredth, offering a precise measure of distance.

Grasping the concept of square roots is integral to solving problems in geometry and algebra, ensuring accuracy in results and enhancing computational proficiency. Understanding how to handle square roots helps in navigating through complex mathematical tasks with assurance.