Problem 100 - College Algebra

Question

Will help you prepare for the material covered in the next section. Let \(\left(x_{1}, y_{1}\right)=(7,2) \quad\) 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 points (7, 2) and (1, -1) is \(3\sqrt{5}\)

Step 1: Identify the Given Points

The points given are \((x_1, y_1) = (7, 2)\) and \((x_2, y_2) = (1, -1)\).

Step 2: Substitute the Coordinates into the Formula

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

Step 3: Simplify the Expression Inside the Square Root

Simplify the expression inside the square root. That results in \(\sqrt{(-6)^2 + (-3)^2} = \sqrt{36 + 9}\).

Step 4: Calculate the Value Inside the Square Root

Calculate the value inside the square root to get \(\sqrt{45}\)

Step 5: Express the Answer in Simplified Radical Form

45 can be expressed as \(9*5\), where 9 has a square root. So, the simplified radical form of \(\sqrt{45}\) is \(3\sqrt{5}\)

Key Concepts

Simplified Radical Form

Understanding the simplified radical form is crucial when working with the distance formula or any algebraic expression involving square roots. This form requires expressing the square root of a number in the most reduced form. To do this, one looks for perfect square factors of the number under the radical sign. For instance, in our problem, we have \( \sqrt{45} \) which can be simplified since 45 has the perfect square factor of 9 (\(9 \times 5\)). Taking the square root of this perfect square, we obtain 3, and we're left with the simpler and more digestible expression \( 3\sqrt{5} \), which is the simplified radical form.

Let's consider another example for better comprehension: if we have a number like 72 under the square root, it can be broken down into \( 36 \times 2 \) because 36 is a perfect square. Thus, \( \sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2} \), which is the simplified radical form of \( \sqrt{72} \). Keeping expressions in this form, especially when dealing with roots, is not only neater but also essential for further algebraic manipulation.

Coordinate Geometry

Coordinate geometry, sometimes known as analytic geometry, is a branch of geometry that describes the link between algebra and geometry through the use of an ordered pair of numbers known as coordinates. In the exercise, we're given two points in a two-dimensional space: \( (x_1, y_1) = (7, 2) \) and \( (x_2, y_2) = (1, -1) \). These points can be plotted on a Cartesian plane, with the first number representing the horizontal position (x-coordinate) and the second number representing the vertical position (y-coordinate).

The distance formula \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) is rooted in the Pythagorean theorem. It's used to calculate the straight-line distance between two points in this plane. Coordinate geometry not only helps us find distances but is also foundational for understanding shapes, slopes, and equations of lines, by translating them into algebraic expressions and equations. This dual perspective makes it a powerful tool in mathematics and many real-world applications.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operations that together represent a specific value or range of values. In our distance formula example, \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) is an algebraic expression that mathematically encodes the distance between two points in terms of their coordinates.

Breaking it down, this expression includes subtraction (\( x_2 - x_1 \) and \( y_2 - y_1 \)), exponentiation (squaring the differences), addition (adding the squares), and taking the square root of the sum. Each of these operations transforms the original coordinates into a meaningful number: the distance. Algebraic expressions are versatile and can represent virtually any numerical relationship, making them central to not only algebra but all branches of mathematics. Simplifying and manipulating these expressions are key skills for students, allowing them to find solutions to a wide array of problems.