Problem 18 - College Algebra

Question

Find the distance between each pair of points. If necessary, round answers to two decimals places. $$ \left(-\frac{1}{4},-\frac{1}{7}\right) \text { and }\left(\frac{3}{4}, \frac{6}{7}\right) $$

Step-by-Step Solution

Answer

The distance between the points (-1/4, -1/7) and (3/4, 6/7) is approximately 1.41.

Step 1: Identify the Coordinates

Identify the X and Y coordinates of the two points. The first point is (-1/4, -1/7) and the second point is (3/4, 6/7).

Step 2: Apply the Distance Formula

Substitute these values into the distance formula. Here, x1 = -1/4, y1 = -1/7, x2 = 3/4, y2 = 6/7. \[Distance = \sqrt{((3/4 - (-1/4))^2 + (6/7 - (-1/7))^2)}\]

Step 3: Simplify the Expression

Simplify the expression inside the square root. \[Distance = \sqrt{(1)^2 + (1)^2} = \sqrt{2}\]

Step 4: Round to Two Decimal Places

Lastly, round the square root of 2 to two decimal places. Using a calculator, √2 comes out approximately as 1.41

Key Concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is a powerful branch of mathematics that combines algebra and geometry to allow for the precise analysis of geometric shapes, using a coordinate system. Typically, the coordinate system used is the Cartesian plane, where points are defined by an ordered pair of numbers \( (x, y) \), often representing their positions along the horizontal (x-axis) and vertical (y-axis) dimensions respectively.

When working with coordinate geometry, the location of any point can be identified by these two values relative to the origin \( (0, 0) \), which is the center point of the plane. This system makes it possible to perform algebraic calculations to find distances, midpoints, gradients, and equations of geometric figures. For example, the distance between two points can be precisely determined, as seen in the exercise where two points are given and the distance between them is queried.

College Algebra

In the realm of college algebra, one focuses on an array of functions, equations, and the comprehensive understanding of algebraic principles. This foundational subject helps establish the groundwork for more advanced mathematical concepts and applications. It is here that students learn to manipulate algebraic expressions, solve equations, and understand functions. This preparation is crucial when diving into problems like calculating distances using the distance formula—a tool derived directly from algebraic operations.

The process of identifying and plugging coordinates into the distance formula can be traced back to algebraic principles like simplifying expressions and square roots. As in the example provided, simplifying the expression inside the square root and rounding off are skills honed through college algebra, illustrating the real-world application of algebraic concepts.

Pythagorean Theorem

At the heart of the distance formula lies the Pythagorean Theorem, a principle stemming from the robust field of geometry. Named after the ancient Greek mathematician Pythagoras, the theorem asserts that in a right-angled triangle, the square of the length of the hypotenuse \(c\) is equal to the sum of the squares of the other two sides \(a\) and \(b\): \[c^2 = a^2 + b^2\].

When we apply this theorem to coordinate geometry, it serves as the foundation for deriving the distance between any two points \( (x_1, y_1) \) and \( (x_2, y_2) \). The two points can be imagined as forming a right-angled triangle with the distance between them as the hypotenuse. Thus, the distance formula \[Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\] reflects the Pythagorean Theorem. By visualizing geometric concepts algebraically, students can navigate between symbols and shapes, affirming the interconnectedness of mathematics.