Problem 96 - Intermediate Algebra

Question

Find the midpoint of the line segment whose endpoints are given. See Example 7 $$ (\sqrt{8},-\sqrt{12}),(3 \sqrt{2}, 7 \sqrt{3}) $$

Step-by-Step Solution

Answer

The midpoint is \( \left( \frac{5\sqrt{2}}{2}, \frac{5\sqrt{3}}{2} \right) \).

Step 1: Identify the coordinates

First, identify the coordinates of the endpoints given in the problem. The coordinates of the first endpoint are \( (\sqrt{8}, -\sqrt{12}) \), and the coordinates of the second endpoint are \( (3 \sqrt{2}, 7 \sqrt{3}) \).

Step 2: Simplify the coordinates

Simplify the square roots in the coordinates. \( \sqrt{8} = 2\sqrt{2} \) and \( -\sqrt{12} = -2\sqrt{3} \). So, the first point is \( (2\sqrt{2}, -2\sqrt{3}) \). The second point \( (3 \sqrt{2}, 7 \sqrt{3}) \) doesn't need simplification.

Step 3: Use the midpoint formula

Use the midpoint formula for a line segment with endpoints \((x_1, y_1)\) and \((x_2, y_2)\), which is given by \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).

Step 4: Calculate the midpoint coordinates

Plug the simplified coordinates into the midpoint formula: \[\left( \frac{2\sqrt{2} + 3\sqrt{2}}{2}, \frac{-2\sqrt{3} + 7\sqrt{3}}{2} \right).\]

Step 5: Perform the arithmetic operations

Calculate each component separately: The \(x\)-component is \( \frac{2\sqrt{2} + 3\sqrt{2}}{2} = \frac{5\sqrt{2}}{2} \). The \(y\)-component is \( \frac{-2\sqrt{3} + 7\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} \). So, the midpoint is \( \left( \frac{5\sqrt{2}}{2}, \frac{5\sqrt{3}}{2} \right) \).

Key Concepts

Line Segment

A line segment is a part of a line that has two distinct endpoints. It is the simplest shape you can understand within the realm of geometry. Think of it as the shortest path connecting two points. **Properties of Line Segments:**

• Has a fixed length, as it doesn't stretch indefinitely like a line does.
• Can reside in one, two, or three dimensions, but in coordinate geometry, it commonly lies in a two-dimensional (2D) plane.

When we talk about finding the midpoint of a line segment, it means that we are determining the exact middle point between its two endpoints. Imagine cutting the line segment into two equal parts; the point where you make the cut is the midpoint. In math, especially in problems like the one provided, understanding the line segment helps in visualizing the role of coordinates and the utilization of formulas to find significant points such as the midpoint.

Coordinate Geometry

Coordinate geometry, often referred to as analytic geometry, combines algebra and geometry to establish relationships between points, lines, and their corresponding coordinates in a plane. It is essential for interpreting geometric figures with precision in terms of numbers.**Basics of Coordinate Geometry:**

• Uses ordered pairs (x, y) to define the position of points within a grid or coordinate plane.
• The horizontal line is called the x-axis, while the vertical line is the y-axis.
• The point where both axes meet is the origin, designated as (0, 0).

In the context of finding a midpoint, coordinate geometry helps break down the task of locating a central point into a straightforward calculation. By using the formula: \[\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\]...you average the x-coordinates and the y-coordinates of the endpoints to find the midpoint. This mathematical process allows you to determine an exact middle point with ease and accuracy.

Square Roots

Square roots are fundamental concepts in mathematics used to find a number that, when multiplied by itself, yields the original number.**Understanding Square Roots:**

• Expressed symbolically with the radical sign \( \sqrt{} \).
• For example, the square root of 9 is 3, because 3 × 3 equals 9.
• While calculating the square root of a non-perfect square, the result is an irrational number, like \( \sqrt{2} \).

In problems like the one we discussed, square roots may appear within coordinates. Simplifying these expressions is essential before plugging them into formulas. For example, simplifying \( \sqrt{8} \) to \( 2\sqrt{2} \) makes calculations manageable.Real-world applications often call for simplifying or estimating square roots to keep numbers practical and solutions concise. In coordinate geometry, handling these roots becomes a frequent task, sharpening your skills in both algebra and arithmetic.