Problem 16 - College Algebra

Question

Find the distance between each pair of points. If necessary, round answers to two decimals places. $$(2 \sqrt{3}, \sqrt{6})\( and \)(-\sqrt{3}, 5 \sqrt{6})$$

Step-by-Step Solution

Answer

The distance between the points \((2 \sqrt{3}, \sqrt{6})\) and \((-\sqrt{3}, 5 \sqrt{6})\) is approximately 11.09.

Step 1: Write down the distance formula

The formula for the distance between two points \((x1,y1)\) and \((x2,y2)\) in a plane is \(d = \sqrt{{(x2 - x1)^2 + (y2 - y1)^2}}\)

Step 2: Substitute the given points into the formula

Substitute the coordinates of the given points into the formula. So, \(x1 = 2 \sqrt{3}, y1 = \sqrt{6}, x2 = -\sqrt{3}, y2 = 5 \sqrt{6}\). Thus, the formula becomes \(d = \sqrt{{(-\sqrt{3} - 2 \sqrt{3})^2 + (5 \sqrt{6} - \sqrt{6})^2}}\) which simplifies to \(d = \sqrt{{(-3 \sqrt{3})^2 + (4 \sqrt{6})^2}}\)

Step 3: Calculate the squares

Calculate the squares in the formula. This yields \(d = \sqrt{{(-3 \sqrt{3})^2 + (4 \sqrt{6})^2}} = \sqrt{{27 + 96}}\)

Step 4: Calculate the square root

Calculate the square root of the sum which is the square root of 123. So, \(d = \sqrt{{123}}\)

Step 5: Round to two decimal places

If rounding is necessary, use a calculator to find the value and round to two decimal places. Hence, \(d \approx 11.09\)

Key Concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is a branch of mathematics that connects algebra and geometry through the use of coordinates. In a two-dimensional plane, any point is identified by an ordered pair \(x, y\). These coordinates signify the location of a point by its horizontal (x) and vertical (y) distances from a reference point, usually called the origin \(0, 0\). Coordinate geometry allows us to explore geometric concepts using algebraic equations.

In the given problem, we are working with the points \((2 \sqrt{3}, \sqrt{6})\) and \((-\sqrt{3}, 5 \sqrt{6})\). Each of these points has an x-coordinate and a y-coordinate. By understanding how to plot these on a graph, you gain insight into the spatial relationships and distances between them. This leads us to the next concept: calculating the distance.

Distance Calculation

The distance formula is a vital tool in coordinate geometry, helping us find the distance between two points on the plane. Derived from the Pythagorean theorem, the formula computes the straight-line distance between two points by measuring the change in x-coordinates and y-coordinates.

The formula is:

• \( d = \sqrt{{(x2 - x1)^2 + (y2 - y1)^2}} \)

By substituting the coordinates from the problem:

• \((x1, y1) = (2 \sqrt{3}, \sqrt{6})\)
• \((x2, y2) = (-\sqrt{3}, 5 \sqrt{6})\)

We plug them into the formula:

• \(d = \sqrt{{(-\sqrt{3} - 2 \sqrt{3})^2 + (5 \sqrt{6} - \sqrt{6})^2}}\)

This simplifies the complex expressions into a manageable format for calculation.

Square Root Simplification

Once we substitute the coordinates into the distance formula, the next step is simplifying the square root expression. Here, you resolve the differences between coordinates and square them:

• Calculate \((-\sqrt{3} - 2 \sqrt{3})^2\): This results in \((-3 \sqrt{3})^2 = 27\).
• Calculate \((5 \sqrt{6} - \sqrt{6})^2\): This yields \( (4 \sqrt{6})^2 = 96\).

Add these squared terms together:

• \(d = \sqrt{27 + 96} = \sqrt{123}\)

By calculating the square root of 123, we bring it to a simpler form for further rounding. This step illustrates how understanding algebraic principles aids in mathematical simplification.

Rounding Decimals

The final step is rounding the calculated distance to two decimal places, which makes the answer more digestible and practical, especially when precise measurements aren't necessary. This involves using a calculator to compute the square root:

• \(\sqrt{123} \) evaluates approximately to 11.09057628.

By rounding to two decimal places, we simplify this to approximately 11.09. Rounding provides a way to express your results in a way that is clear and concise, maintaining accuracy appropriate for many real-world applications.

Remember:

• If the third decimal is 5 or more, round up.
• If it’s less than 5, round down.

This practice of rounding helps us present answers cleanly while preserving the needed precision.