Problem 106 - A Graphical Approach to Precalculus with Limits

Question

Prove that the midpoint \(M\) of the line segment joining endpoints \(P\left(x_{1}, y_{1}\right)\) and \(Q\left(x_{2}, y_{2}\right)\) has coordinates $$ \left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) $$ by showing that the distance between \(P\) and \(M\) is equal to the distance between \(M\) and \(Q\) and that the sum of these distances is equal to the distance between \(P\) and \(Q\).

Step-by-Step Solution

Answer

Coordinates of midpoint M are correctly given by \( \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \).

Step 1: Find the coordinates of the midpoint M

To find the coordinates of the midpoint \( M \) of a line segment connecting two endpoints \( P(x_1, y_1) \) and \( Q(x_2, y_2) \), we calculate the averages of the x-coordinates and y-coordinates of the endpoints. This results in the coordinates \( M \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).

Step 2: Calculate the distance from P to M

Using the distance formula, the distance from \( P(x_1, y_1) \) to \( M \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \) is \[ D_{PM} = \sqrt{\left( \frac{x_1 + x_2}{2} - x_1 \right)^2 + \left( \frac{y_1 + y_2}{2} - y_1 \right)^2 } \]Simplifying, this becomes \( D_{PM} = \frac{1}{2} \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).

Step 3: Calculate the distance from M to Q

Similarly, calculate the distance from \( M \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \) to \( Q(x_2, y_2) \) using the distance formula:\[D_{MQ} = \sqrt{\left( x_2 - \frac{x_1 + x_2}{2} \right)^2 + \left( y_2 - \frac{y_1 + y_2}{2} \right)^2 }\] Simplifying this yields \( D_{MQ} = \frac{1}{2} \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).

Step 4: Verify that distances PM and MQ are equal

From our calculations in Step 2 and Step 3, we found that \( D_{PM} = \frac{1}{2} \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) and \( D_{MQ} = \frac{1}{2} \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). Hence, \( D_{PM} = D_{MQ} \).

Step 5: Sum of distances PM and MQ equals distance PQ

The total distance \( PQ \) between points \( P \) and \( Q \) is given by the distance formula as:\[ D_{PQ} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]The sum of \( D_{PM} \) and \( D_{MQ} \) is:\[\frac{1}{2} \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} + \frac{1}{2} \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\] Thus, \( D_{PM} + D_{MQ} = D_{PQ} \). This verifies that \( M \) is indeed the midpoint.

Key Concepts

Distance Formula

The distance formula is a crucial tool in coordinate geometry. It helps us to calculate the distance between two points on a coordinate plane. The formula originates from the Pythagorean theorem and is given by:\[ D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]Where \( (x_1, y_1) \) and \( (x_2, y_2) \) are the coordinates of the two points.

• Measure the horizontal distance by subtracting the x-coordinates: \( x_2 - x_1 \)
• Measure the vertical distance by subtracting the y-coordinates: \( y_2 - y_1 \)

Then, square each of these distances. Sum them up, and take the square root to find the distance. This formula is not only important for calculating straight-line distances but is also foundational for solving more complex geometric problems. In the context of the exercise, we use it to check that the distances from a midpoint to each endpoint of a line segment are equal.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is the study of geometry using a coordinate system. The most common coordinate system used is the Cartesian coordinate system, where any point on the plane can be represented by an ordered pair \( (x, y) \).This branch of geometry bridges algebra and geometry by describing geometric figures algebraically. Points, lines, and circles can all be represented with algebraic equations, allowing us to perform calculations and solve problems that would be challenging to visualize or solve geometrically.

• Enables converting geometric problems into algebraic equations
• Helps in determining lengths, midpoints, and areas using algebraic methods

In the given problem, coordinate geometry allows us to algebraically find the midpoint of the line segment using the coordinates of its endpoints. This method is not only precise but also demonstrates why the midpoint formula works by verifying distance properties.

Line Segment

A line segment is a part of a line that is bounded by two distinct endpoints. Unlike a line, it does not extend infinitely. You can measure a line segment's length using the distance formula. It is essentially the straight path connecting two points in a plane or space.The concept of a line segment is crucial because it forms the basis for defining other geometric constructs such as polygons and polyhedra. In geometry, understanding how to navigate and measure these segments is important for solving a variety of problems.

• A line segment has a finite length, which can be calculated
• It is defined by its two endpoints
• The midpoint divides it into two equal segments

In the exercise on midpoints, the line segment connects points \( P(x_1, y_1) \) and \( Q(x_2, y_2) \). Finding the midpoint \( M \) involves calculating the average of the x-coordinates and y-coordinates of the endpoints. This precise division into two equal halves is what makes the midpoint so important and useful in geometry.