Problem 44 - College Algebra with Modeling and Visualization

Question

Find the midpoint of the line segment connecting the points. $$ (\sqrt{7}, 3 \sqrt{3}),(-\sqrt{7},-\sqrt{3}) $$

Step-by-Step Solution

Answer

The midpoint is \\((0, \\sqrt{3})\\).

Step 1: Understanding the Concept

The midpoint of a line segment connecting two points \( (x_1, y_1) \, ext{and} \, (x_2, y_2) \) is calculated using the formula \( ext{Midpoint} = rac{(x_1 + x_2)}{2}, rac{(y_1 + y_2)}{2} \). This gives the point that lies exactly halfway between the two points.

Step 2: Identifying the Coordinates

The coordinates of the given points are \( (x_1, y_1) = (\sqrt{7}, 3 \sqrt{3}) \) and \( (x_2, y_2) = (-\sqrt{7}, -\sqrt{3}) \). These values will be used in the midpoint formula.

Step 3: Calculating the Midpoint's x-coordinate

Substitute \( x_1 = \sqrt{7} \) and \( x_2 = -\sqrt{7} \) into the formula for the x-coordinate: \[\frac{\sqrt{7} + (-\sqrt{7})}{2} = \frac{0}{2} = 0\]

Step 4: Calculating the Midpoint's y-coordinate

Substitute \( y_1 = 3 \sqrt{3} \) and \( y_2 = -\sqrt{3} \) into the formula for the y-coordinate: \[\frac{3 \sqrt{3} + (-\sqrt{3})}{2} = \frac{2 \sqrt{3}}{2} = \sqrt{3}\]

Step 5: Combining the Midpoint Coordinates

Combine the x-coordinate and y-coordinate calculated in the previous steps to determine the midpoint: \( (0, \sqrt{3}) \). This is the midpoint of the line segment.

Key Concepts

Coordinate Geometry

Coordinate geometry, or analytic geometry, is the study of geometry using a coordinate system. This powerful mathematical tool allows us to represent geometric shapes in a numerical way. By using points defined by coordinates \( (x, y) \), we can perform a variety of operations and solve problems that involve geometric figures.
In the exercise provided, we are working with points in a two-dimensional plane characterized by coordinates derived from square roots. The significance of coordinate geometry in this context is its ability to provide a clear, visual representation of the problem. By plotting points \( (\sqrt{7}, 3\sqrt{3}) \) and \( (-\sqrt{7}, -\sqrt{3}) \), we can easily visualize the line segment connecting them. From here, tasks like finding the midpoint become more tangible.
Key elements of coordinate geometry include:

• Points: Defined by pairs \( (x, y) \)
• Lines: Connecting two or more points
• Midpoints: The center point of a line segment

By understanding these elements, we can solve problems and understand geometric relationships with ease.

Algebraic Calculation

Algebraic calculations involve manipulating numbers and symbols to find solutions to mathematical problems. In the context of our midpoint problem, we use algebra to calculate the exact center between two complex points.
Calculating a midpoint requires averaging the coordinates of two points. Using the midpoint formula \( \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \), we find the mean of the x-values and the mean of the y-values to obtain the midpoint.
The steps involve:

• Adding the x-values: \( x_1 + x_2 \)
• Dividing by 2 to find the average
• Repeating the process for the y-values

In our problem, after substituting \( x_1 = \sqrt{7} \) and \( x_2 = -\sqrt{7} \), we found the x-coordinate of the midpoint to be 0. Similarly, using \( y_1 = 3\sqrt{3} \) and \( y_2 = -\sqrt{3} \), we calculated the y-coordinate to be \( \sqrt{3} \). The algebraic manipulation of these values highlights the precision and usefulness of algebraic calculations in solving geometry-based problems.

Square Roots

Square roots are fundamental concepts in algebra and geometry, representing a number that gives a specified product when multiplied by itself. In coordinate geometry, square roots can define the coordinates of points, as seen in this exercise. For instance, \( \sqrt{7} \) is one of the x-coordinates, and \( 3\sqrt{3} \) is one of the y-coordinates.
Handling square roots in equations requires precision and understanding of basic root properties. For this solution, it was essential to know:

• \( \sqrt{a} \times \sqrt{a} = a \)
• \( \frac{a + a'}{2} \) deals similarly with root terms like rational numbers

The presence of square roots makes solutions both challenging and interesting. They remind us that coordinate and algebraic operations aren't limited to integers or simple fractions. Calculations involving roots require careful substitution and arithmetic to ensure accuracy. Through practice, understanding how to manipulate these in problems becomes an invaluable skill for students of maths.