Problem 93 - A Graphical Approach to Precalculus with Limits

Question

Find (a) the distance between \(P\) and \(Q\) and (b) the coordinates of the midpoint \(M\) of the segment joining \(P\) and \(Q\). $$P(9.2,3.4), Q(6.2,7.4)$$

Step-by-Step Solution

Answer

The distance is 5 units and the midpoint is (7.7, 5.4).

Step 1: Find the Distance Formula

The distance between two points \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) can be calculated using the formula: \(D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).

Step 2: Substitute the Coordinates

Given the points \(P(9.2, 3.4)\) and \(Q(6.2, 7.4)\), substitute the values into the distance formula:\(D = \sqrt{(6.2 - 9.2)^2 + (7.4 - 3.4)^2}\).

Step 3: Perform the Calculations for Distance

Calculate the differences: \((6.2 - 9.2) = -3\) and \((7.4 - 3.4) = 4\). Then, square the differences: \((-3)^2 = 9\) and \(4^2 = 16\).Finally, sum these values and find the square root:\(D = \sqrt{9 + 16} = \sqrt{25} = 5\).

Step 4: Find the Midpoint Formula

The midpoint \(M(x, y)\) of the segment joining points \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) is given by: \(M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\).

Step 5: Substitute the Coordinates for Midpoint

Using the points \(P(9.2, 3.4)\) and \(Q(6.2, 7.4)\), the midpoint can be calculated as:\(M = \left(\frac{9.2 + 6.2}{2}, \frac{3.4 + 7.4}{2}\right)\).

Step 6: Perform the Calculations for Midpoint

Compute the additions: \(9.2 + 6.2 = 15.4\) and \(3.4 + 7.4 = 10.8\). Now, divide by 2:\(M = \left(\frac{15.4}{2}, \frac{10.8}{2}\right) = (7.7, 5.4)\).

Key Concepts

Midpoint Formula

The midpoint formula is a fundamental concept in coordinate geometry that helps us find the exact center point between two given points. It is represented as follows:

• Midpoint, \(M(x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)

This formula calculates the average of the x-coordinates and y-coordinates separately. Finding the midpoint is useful in various applications like dividing a line segment in half or finding a central location.
For example, with points \(P(9.2, 3.4)\) and \(Q(6.2, 7.4)\):

• Calculate the average of x-coordinates: \(\frac{9.2 + 6.2}{2} = 7.7\)
• Calculate the average of y-coordinates: \(\frac{3.4 + 7.4}{2} = 5.4\)

Thus, the midpoint \(M\) is \((7.7, 5.4)\). With practice, using the midpoint formula becomes an intuitive process to identify the center point in any line segment between two coordinates.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, integrates algebra and geometry using a coordinate plane. It provides a bridge between algebraic equations and geometric figures, allowing for precision in calculations and analysis of shapes.
In coordinate geometry, points are represented with pairs of numbers (x, y) on a two-dimensional plane. Lines and curves then form based on those points. This system enables us to:

• Determine distances between points.
• Find midpoints and slopes of lines.
• Analyze the properties of geometric shapes.

For instance, using coordinate geometry, we can accurately describe the location of any point and solve problems like finding the shortest distance between two locations or determining the slope of a hill. The principles of coordinate geometry are essential in many real-life applications including navigation, computer graphics, and architectural design.

Distance Calculation

Distance calculation is an essential component of understanding relationships between points in coordinate geometry. The distance formula helps in determining the straight-line distance between any two points on a plane.
The formula is derived from the Pythagorean theorem and is expressed as:

• \(D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

This formula measures the hypotenuse of a right triangle formed by the difference in x and y-coordinates. Calculating distance becomes straightforward by substituting the coordinates of the two points into the formula.
For example, to calculate the distance between points \(P(9.2, 3.4)\) and \(Q(6.2, 7.4)\):

• Calculate the differences: \((6.2 - 9.2) = -3\) and \((7.4 - 3.4) = 4\)
• Square these differences: \((-3)^2 = 9\) and \(4^2 = 16\)
• Sum the squares and take the square root: \(D = \sqrt{9 + 16} = \sqrt{25} = 5\)

Hence, the distance \(D\) between points \(P\) and \(Q\) is 5 units. Mastering distance calculation empowers students to solve complex geometry problems efficiently.