Problem 31 - Precalculus

Question

Suppose that the coordinates of points \(P, Q,\) and \(M\) are $$\begin{array}{c}P\left(x_{1}, y_{1}\right) \quad Q\left(x_{2}, y_{2}\right) \\\M\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\end{array}$$ Follow steps (a) and (b) to prove that \(M\) is the midpoint of the line segment from \(P\) to \(Q\) (a) By computing both of the distances \(P M\) and \(M Q\) show that \(P M=M Q\). (This shows that \(M\) lies on the perpendicular bisector of line segment \(\overline{P Q}\), but it does not show that \(M\) actually lies on \(P Q .\) ) (b) Show that \(P M+M Q=P Q\). (This shows that \(M\) does lie on \(\overline{P Q} .)\)

Step-by-Step Solution

Answer

Point M is the midpoint of the line segment PQ.

Step 1: Calculate Distance PM

To calculate the distance between points \(P\) and \(M\), we use the distance formula: \[PM = \sqrt{\left(\frac{x_1+x_2}{2} - x_1\right)^2 + \left(\frac{y_1+y_2}{2} - y_1\right)^2}\]Simplifying the expressions inside the parentheses:\[PM = \sqrt{\left(\frac{x_2-x_1}{2}\right)^2 + \left(\frac{y_2-y_1}{2}\right)^2}\]So we have:\[PM = \frac{1}{2}\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\]

Step 2: Calculate Distance MQ

Similarly, to calculate the distance between points \(M\) and \(Q\), we use the distance formula:\[MQ = \sqrt{\left(x_2 - \frac{x_1+x_2}{2}\right)^2 + \left(y_2 - \frac{y_1+y_2}{2}\right)^2}\]Simplifying:\[MQ = \sqrt{\left(\frac{x_2-x_1}{2}\right)^2 + \left(\frac{y_2-y_1}{2}\right)^2}\]This results in:\[MQ = \frac{1}{2}\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\]

Step 3: Verify that PM = MQ

Since both \(PM\) and \(MQ\) are: \[\frac{1}{2}\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\]it follows that \(PM = MQ\).

Step 4: Calculate Distance PQ

Calculate the distance from \(P\) to \(Q\):\[PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\]

Step 5: Verify that PM + MQ = PQ

We have already calculated that both \(PM\) and \(MQ\) are:\[PM = MQ = \frac{1}{2}\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\]Thus:\[PM + MQ = \frac{1}{2}\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} + \frac{1}{2}\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = PQ\]

Step 6: Conclude that M is the midpoint

Since \(PM = MQ\) and \(PM + MQ = PQ\), point \(M\) is equidistant from \(P\) and \(Q\), and \(M\) lies on \(\overline{PQ}\). Therefore, \(M\) is the midpoint of \(\overline{PQ}\).

Key Concepts

Distance Formula

The Distance Formula is a valuable tool in coordinate geometry for determining the distance between two points in a plane. This formula arises from the Pythagorean theorem and is expressed as follows: for two points \( (x_1, y_1) \) and \( (x_2, y_2) \), the distance \( d \) between them is given by:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]The formula calculates the straight-line distance, known as the Euclidean distance, between the two points.
This direct application of the Pythagorean theorem helps in various geometrical problems, such as determining the length of a line segment, finding the radius of a circle with a given center, or verifying geometric properties of figures.

• It is important to correctly substitute the coordinates into the formula.
• Ensure that the computations inside the square root are simplified accurately.

In our exercise, we used this formula to calculate the distances \( PM \), \( MQ \), and \( PQ \), allowing us to deduce properties of the points \( M \), \( P \), and \( Q \) in the context of a line segment.

Midpoint Theorem

The Midpoint Theorem states that the midpoint of a line segment is the point that divides the segment into two equal halves. If you have a line segment connecting points \( P(x_1, y_1) \) and \( Q(x_2, y_2) \), the coordinates of the midpoint \( M \) can be determined using the following formula:\[M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\]This formula calculates the arithmetic mean of the coordinates, effectively finding the "center" of the line segment.
In context, the midpoint has several applications in geometry and related fields:

• Determining whether a point is centrally located within a given segment.
• Helps in finding lines or shapes' symmetrical properties.

In the exercise example, to prove that \( M \) is indeed the midpoint of \( \overline{PQ} \), we first calculated that \( M \) is equidistant from both \( P \) and \( Q \). Thus, satisfying the conditions set by the Midpoint Theorem, indicating \( M \)'s position as the center.

Line Segment

A line segment is one of the fundamental concepts in geometry, representing the shortest path between two points. It connects two endpoints, containing all the points in between based on the straight line principle. In coordinate geometry, line segments can be analyzed using their endpoints to determine various attributes, such as length and midpoint.

• The length of a line segment is calculated using the Distance Formula.
• The midpoint can be determined using the Midpoint Theorem.

Understanding these properties is critical in solving complex geometric problems.
In our exercise, using the endpoints \( P \) and \( Q \), we applied these formulas to demonstrate the center point \( M \) is equidistant, thereby classifying \( M \) as the midpoint of the segment \( \overline{PQ} \). This understanding is crucial in establishing symmetry and balance within geometrical shapes and forms.