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)\( and \)\left(\frac{3}{4}, \frac{6}{7}\right)$$

Step-by-Step Solution

Answer

The distance between the two points is approximately \(1.41\)

Step 1: Identify the coordinates

The coordinates of the first point are \(-\frac{1}{4}\) and \(-\frac{1}{7}\). The coordinates of the second point are \(\frac{3}{4}\) and \(\frac{6}{7}\). We will refer to these as \(x_1, y_1\) for the first point, and \(x_2, y_2\) for the second point respectively.

Step 2: Substitute into formula

Substitute the coordinates into the distance formula: \(d=\sqrt{(\frac{3}{4}-(-\frac{1}{4}))^2 + (\frac{6}{7}-(-\frac{1}{7}))^2}\).

Step 3: Simplify

Simplify within the square root: \(d=\sqrt{(\frac{4}{4})^2 + (\frac{7}{7})^2}=\sqrt{1^2 + 1^2}=\sqrt{1+1}\).

Step 4: Result

So the distance \(d=\sqrt{2}\), which is approximately \(1.41\) when rounded off to two decimal places.

Key Concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is a branch of mathematics that uses algebraic equations to describe and analyze points, lines, and shapes on a coordinate plane. A coordinate plane is a two-dimensional surface formed by two intersecting lines: the horizontal axis known as the x-axis, and the vertical axis known as the y-axis. Each point on this plane can be specified by an ordered pair of numbers \( (x, y) \) where \( x \) represents the horizontal position, and \( y \) represents the vertical position.

For example, in the given exercise, the points \(\left(-\frac{1}{4},-\frac{1}{7}\right)\) and \(\left(\frac{3}{4}, \frac{6}{7}\right)\) represent specific locations on the coordinate plane. To find the distance between any two points in coordinate geometry, we use the distance formula, which is derived from the Pythagorean theorem and operates on the principle that the distance between points is the hypotenuse of a right-angled triangle formed by these points.

Simplifying Square Roots

When working with square roots, it is often necessary to simplify them in order to better understand and work with the numerical values. Simplification usually involves turning the square root of a non-square number into a more manageable form, often by factoring out perfect squares or rationalizing denominators.

In the context of the exercise, after applying the distance formula, we end up with the square root of a sum of squares. The expression under the square root \(\sqrt{1^2 + 1^2} = \sqrt{1+1}\) needs to be simplified. Since the numbers are already perfect squares, we simply add them inside the radical to get \(\sqrt{2}\). While \(\sqrt{2}\) itself is an irrational number and cannot be further simplified as an exact value, we often round it for practical purposes. In this case, we round \(\sqrt{2}\) to 1.41, which is an approximation to two decimal places. Understanding how to simplify square roots is essential in many fields of mathematics including algebra, geometry, and calculus.

Pythagorean Theorem

The Pythagorean theorem is one of the most well-known principles in mathematics, fundamental in the study of Euclidean geometry. It states 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 \( b \): \( c^2 = a^2 + b^2 \).

This theorem is crucial when we want to find the distance between two points in the Cartesian coordinate system, as the distance formula is an application of the Pythagorean theorem. In our exercise, the distance between the points is effectively the length of the hypotenuse in a right-angled triangle, with the horizontal and vertical distances between the points serving as the other two sides. By applying the theorem to these distances, which we calculate by finding the differences in the x and y coordinates, we obtain the square of the distance, which is then square rooted to provide the actual distance.