Problem 85

Question

Find the distance between each pair of points. Give an exact distance and a three-decimal-place approximation. (1.7,-3.6) and (-8.6,5.7)

Step-by-Step Solution

Verified
Answer
The exact distance is \(\sqrt{192.58}\). The approximate distance is 13.876.
1Step 1: Identify the Points
The given points are \((x_1, y_1) = (1.7, -3.6)\) and \((x_2, y_2) = (-8.6, 5.7)\).
2Step 2: Apply Distance Formula
Use the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Substitute the coordinates of the points into this formula.
3Step 3: Calculate Horizontal and Vertical Differences
Calculate the difference between the x-coordinates: \(-8.6 - 1.7 = -10.3\).Calculate the difference between the y-coordinates: \(5.7 - (-3.6) = 9.3\).
4Step 4: Square the Differences
Square the horizontal difference: \((-10.3)^2 = 106.09\).Square the vertical difference: \((9.3)^2 = 86.49\).
5Step 5: Add the Squared Differences
Add the squared differences: \(106.09 + 86.49 = 192.58\).
6Step 6: Compute the Square Root
Find the square root of the sum of the squares: \(\sqrt{192.58}\) to obtain the exact distance.
7Step 7: Approximate Decimal Value
Compute \(\sqrt{192.58}\) to three decimal places, which gives approximately 13.876.

Key Concepts

Coordinate GeometryPoint Distance CalculationExact and Approximate Distances
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, is a branch of mathematics that utilizes algebraic equations to denote geometric shapes and solve geometric problems using a coordinate plane. A coordinate plane consists of two perpendicular axes, typically labeled as the x-axis (horizontal) and y-axis (vertical). Each point on this plane is identified by a pair of numerical values called coordinates. The coordinates are usually written as \((x, y)\), where 'x' signifies the position along the horizontal axis and 'y' indicates the position along the vertical axis.

In the context of coordinate geometry, finding distances, such as between two points, involves calculating how far apart these points are on the plane. This concept is crucial for various applications, including engineering, computer graphics, and navigation. It simplifies the process of crossing from elementary arithmetic and geometry to more complex algebraic and geometric analyses. By understanding the coordinate geometry framework, students can better understand spatial relationships and computations.
Point Distance Calculation
The distance between two points on a coordinate plane can be calculated using the distance formula derived from the Pythagorean theorem. Let's break it down step-by-step:
  • Firstly, identify the coordinates of the two points, \((x_1, y_1)\) and \((x_2, y_2)\).
  • The distance formula is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
This formula calculates the straight-line distance between two points by considering these points as vertices of a right triangle.

The horizontal leg is the difference between the x-coordinates \((x_2 - x_1)\) and the vertical leg is the difference between the y-coordinates \((y_2 - y_1)\). Once these differences are squared and summed, taking the square root of this sum gives the distance between the two points. This powerful tool efficiently measures the space between any two locations on the plane.
Exact and Approximate Distances
Understanding exact and approximate distances is essential when calculating the length between two points in a coordinate plane.

An exact distance is the most precise measurement, usually expressed as a radical expression (i.e., in terms of square roots that aren't simplified to decimals). This is because many times, like in our exercise, these calculated square roots produce irrational numbers, which have an infinite and non-repeating decimal expansion. For example, if \(\sqrt{192.58}=\)"an exact value in radical form," it might be left in this form for mathematical accuracy.

On the other hand, approximate distances are derived by computing the exact distance to a specific decimal place, usually when a numeric value is easier to work with or required for practical purposes. In our exercise, \(\sqrt{192.58}\)approximately equals 13.876 when rounded to three decimal places. This approximation helps in scenarios where using a precise decimal is needed, such as in applications requiring concise or interval valuations.