Problem 100 - College Algebra Essentials

Question

Exercises \(100-102\) will help you prepare for the material covered in the next section. Let \(\quad\left(x_{1}, y_{1}\right)=(7,2) \quad\) and \(\quad\left(x_{2}, y_{2}\right)=(1,-1) . \quad\) Find \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} .\) Express the answer in simplified radical form.

Step-by-Step Solution

Answer

The distance between the two points is \(3\sqrt{5}\) units

Step 1: Identify the Given Points

Given two points in the form (x, y) - point 1 is (7, 2) and point 2 is (1, -1)

Step 2: Implement the Distance Formula

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

Step 3: Substitute the given points into the formula

Substitute the coordinates of the given points into the distance formula: \(\sqrt{(1 - 7)^2 + (-1 - 2)^2}\)

Step 4: Perform the calculations

Calculate the individual squared differences, then their sum, and lastly take the square root of that sum to get the distance: \(\sqrt{(-6)^2 + (-3)^2} = \sqrt{36 + 9}= \sqrt{45}\)

Step 5: Simplify the radical

To simplify \(\sqrt{45}\), find the prime factorization of 45 which is \(3 * 3 * 5\), then take out in pairs. This gives \(3\sqrt{5}\)

Key Concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, merges algebra and geometry using a coordinate system to solve geometry problems. It allows us to calculate distances, angles, and other properties, by placing geometric figures within a coordinate system and using formulas, especially useful when working with planes.

• In coordinate geometry, every point is defined by a set of coordinates \((x, y)\).
• You can find geometric relationships like midpoints, slopes, and distances between points algebraically.

By assigning coordinates to geometric shapes, complex geometric problems become simple arithmetic tasks. In this exercise, we are dealing with a standard application of coordinate geometry: calculating the distance between two points.

Simplified Radical Form

When dealing with expressions that involve square roots, it’s often best to simplify them into their most reduced or simplified form. This process ensures the expression is in a format that is easiest to read or use in further calculations.

• Simplifying a square root involves factoring the number under the square root into its prime factors.
• You then group the factors into pairs, where each pair can become a factor outside the square root.

For instance, given \(\sqrt{45}\), we find its prime factors: \(45 = 3 \times 3 \times 5\). We can simplify this to \(3\sqrt{5}\), because the pair of 3's come out of the square root as a single 3. This technique is critical in making complex calculations more manageable and results more precise.

Distance Between Points

One of the primary uses of coordinate geometry is finding the distance between two points on a plane. The distance formula is derived from the Pythagorean Theorem, and it allows you to calculate this distance using algebra.

• The formula to determine the distance \(d\) between points \((x_1, y_1)\) and \((x_2, y_2)\) is: \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
• This expression calculates the horizontal and vertical distances between the points and then uses these to find the total distance by applying the essence of the Pythagorean Theorem.

In our exercise, substituting the points \((7, 2)\) and \((1, -1)\) into the formula results in \(\sqrt{(-6)^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45}\). This method provides an efficient way to determine the exact distance, which is crucial for accurate measurements in various fields like physics, navigation, and computer graphics.