Problem 54 - Precalculus with Limits

Question

In Exercises 47-56, (a) plot the points, (b) find the distance between the points, and (c) find the midpoint of the line segment joining the points. \( (-\frac{1}{3}, -\frac{1}{3}) \), \( (-\frac{1}{6}, -\frac{1}{2}) \)

Step-by-Step Solution

Answer

The distance between the points is \(\frac{\sqrt{2}}{6}\) and the midpoint is \(-\frac{1}{4}, -\frac{5}{12}\)

Step 1: Plot the Points

The points to plot are \(-\frac{1}{3}, -\frac{1}{3}\) and \(-\frac{1}{6}, -\frac{1}{2}\). These should be clearly marked on a graph in the correct positions relative to each other and the axes.

Step 2: Find the Distance Between the Points

The distance, d, between two points \((x_1, y_1)\) and \((x_2, y_2)\) in a plane is given by the formula: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Substituting our points into this formula, we get \(d = \sqrt{(-\frac{1}{6} + \frac{1}{3})^2 + (-\frac{1}{2} + \frac{1}{3})^2}\) which simplifies to \(d = \sqrt{(\frac{1}{6})^2 + (-\frac{1}{6})^2} = \frac{\sqrt{2}}{6}\).

Step 3: Find the Midpoint

The midpoint, M, between two points \((x_1, y_1)\) and \((x_2, y_2)\) in a plane is given by the formula: \(M = (\frac{x_1 + x_2}{2}, \frac{y_1+ y_2}{2})\). Substituting our points into this formula, we get \(M = (-\frac{1}{3} -\frac{1}{6})/2, (-\frac{1}{3} -\frac{1}{2})/2\), which simplifies to \(M = -\frac{1}{4}, -\frac{5}{12}\).

Key Concepts

Distance Formula

The distance formula is a mathematical equation used to find the length between two points in a coordinate plane. Think of it as the Pythagorean theorem adapted for coordinates. When calculating distance, we treat the segment connecting the two points as the hypotenuse of a right-angled triangle.

Let's take two points, \( (x_1, y_1) \) and \( (x_2, y_2) \). The formula to find the distance \(d\) between these points is \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]. You can see we take the differences in the x-coordinates and y-coordinates, square them, add them up, and then take the square root of the sum.

To provide a real-life analogy, if you had two buildings in a city with coordinates representing their location, the distance formula would help you measure the shortest path a crow would fly between them, bypassing the city's grid layout of streets.

Midpoint Formula

The midpoint formula calculates the exact middle point along the line segment connecting two points in the plane, much like finding the halfway point between two cities on a map. It is given by the formula \[M = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})\].

Here, \(M\) represents the midpoint, and you're essentially averaging the x-coordinates and y-coordinates of the two points to find the center. For example, if you wanted to meet a friend exactly halfway, you’d find the midpoint of your respective locations, ensuring neither of you travels further than the other.

Plotting Points in a Plane

Plotting points in a plane is a foundational skill in coordinate geometry. It involves placing points on a graph based on their x-coordinate (horizontal movement) and y-coordinate (vertical movement).

Here's a simple way to remember how to plot: start at the origin (0,0), move along the x-axis by the value of the x-coordinate, then move parallel to the y-axis by the value of the y-coordinate. Make sure to go right or up for positive values and left or down for negative values. Visualizing points on a graph can help you better understand their relationship with each other and with the axes.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is the study of geometry using a coordinate system. This approach combines algebra and geometry to describe the position, line segments, shapes, and areas of geometric figures on a plane using coordinates - points described by x and y values.

It's the framework that allows us to use algebraic equations to describe geometric shapes and perform calculations like finding distances or midpoints, as in the exercise provided. This powerful tool enables us to solve geometric problems that might be cumbersome or impossible with only classical Euclidean geometry.