Problem 80 - Intermediate Algebra

Question

Find the distance between each pair of points. $$ (\sqrt{8},-\sqrt{20}) \text { and }(\sqrt{50},-\sqrt{45}) $$

Step-by-Step Solution

Answer

The distance is approximately 4.796 units.

Step 1: Identify the Coordinates

We are given two points with coordinates: Point 1: \((\sqrt{8}, -\sqrt{20})\) and Point 2: \((\sqrt{50}, -\sqrt{45})\). Our task is to calculate the distance between these two points.

Step 2: Recall the Distance Formula

The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) in a plane is given by the formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Step 3: Substitute Coordinates into the Formula

Substitute \(x_1 = \sqrt{8}, y_1 = -\sqrt{20}, x_2 = \sqrt{50}, y_2 = -\sqrt{45}\) into the distance formula: \[ d = \sqrt{(\sqrt{50} - \sqrt{8})^2 + (-\sqrt{45} + \sqrt{20})^2} \]

Step 4: Simplify Expressions

Simplify each square root and ensure all calculations are manageable: - Compute each difference inside the parentheses.- For longitudinal clarity, use decimal approximations: \( \sqrt{8} \approx 2.828, \sqrt{20} \approx 4.472, \sqrt{50} \approx 7.071, \sqrt{45} \approx 6.708 \). Substituting these:- \( \sqrt{50} - \sqrt{8} \approx 7.071 - 2.828 = 4.243 \)- \( -\sqrt{45} + \sqrt{20} \approx -6.708 + 4.472 = -2.236 \).

Step 5: Compute Squared Differences

Square both differences obtained:- \((4.243)^2 = 18.000395 \approx 18\)- \((-2.236)^2 = 5.000296 \approx 5\)

Step 6: Calculate the Distance

Add the squared differences and take the square root:\[ d = \sqrt{18 + 5} = \sqrt{23} \approx 4.796 \]

Step 7: Conclusion

The distance between the given points \((\sqrt{8}, -\sqrt{20})\) and \((\sqrt{50}, -\sqrt{45})\) is approximately 4.796 units.

Key Concepts

Coordinate Geometry

Coordinate Geometry is a branch of mathematics that uses a coordinate system to describe and analyze geometrical figures. It represents points on a plane using coordinate pairs, most commonly depicted as \((x, y)\). This allows us to quantify geometric elements such as lines and curves, making complex calculations, such as finding distances or equations of lines more straightforward.
In this exercise, we are tasked with finding the distance between two points: \((\sqrt{8}, -\sqrt{20})\) and \((\sqrt{50}, -\sqrt{45})\). These points, placed on a coordinate plane, represent exact locations defined by numerical values derived from square roots.

• The first point has coordinates \(x_1 = \sqrt{8}\) and \(y_1 = -\sqrt{20}\).
• The second point is given by \(x_2 = \sqrt{50}\) and \(y_2 = -\sqrt{45}\).

Understanding how to locate these points on a plane and how they interact with each other through the use of formulas like the Distance Formula is a key aspect of Coordinate Geometry.

Simplifying Square Roots

Square roots often appear in Coordinate Geometry when calculating precise distances. Simplifying these roots can be crucial for accurate computations and a clearer understanding of the problem.
Let's take a look at how to simplify and calculate square roots as seen in this exercise:

• \(\sqrt{8}\) simplifies to approximately \(2.828\), which is crucial when you need a precise number for calculations.
• \(\sqrt{20}\) becomes roughly \(4.472\), allowing you to work easily in further steps.
• \(\sqrt{50}\) is about \(7.071\), reducing complexity in the Distance Formula.
• \(\sqrt{45}\) simplifies to approximately \(6.708\).

During calculations, it's often necessary to replace square roots with decimal values to perform addition, subtraction, and squaring operations correctly, as indicated in Step 4 of the original solution.
This approach simplifies the application of algebraic formulas and minimizes any potential errors in further calculations.

Intermediate Algebra

Intermediate Algebra bridges the gap between basic algebra and advanced mathematical studies. In the context of this exercise, it involves using algebraic formulas and computations to solve real-world problems.
The Distance Formula is an essential part of Intermediate Algebra, combining several concepts into one application:

• Using expressions such as \((x_2 - x_1)^2\) and \((y_2 - y_1)^2\), which require subtraction and squaring of simplified square roots.
• Handling square roots within the Distance Formula requires an understanding of both arithmetic operations and algebraic expressions.
• Summing individual differences and then taking the square root of that sum is a multi-step computational skill.

In Step 6, adding and then calculating \(\sqrt{23}\) as approximately \(4.796\) demonstrates the importance of these skills.
Proficiency in Intermediate Algebra allows for the effective application of the Distance Formula and simplifies the resolution of complex geometric problems.