Problem 15 - Intermediate Algebra

Question

Verify that the points \((7,12)\) and \((11,18)\) divide the line segment joining \((3,6)\) and \((15,24)\) into three segments of equal length.

Step-by-Step Solution

Answer

Points (7,12) and (11,18) divide the segment into three equal parts.

Step 1: Find the total distance between endpoints

To determine whether the points divide the line segment into three equal parts, we first compute the distance between the endpoints \((3,6)\) and \((15,24)\). The distance formula is \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).For our points, \(x_1 = 3, y_1 = 6, x_2 = 15, y_2 = 24\). Thus:\[ d = \sqrt{(15 - 3)^2 + (24 - 6)^2} = \sqrt{12^2 + 18^2} = \sqrt{144 + 324} = \sqrt{468} = 6\sqrt{13} \]

Step 2: Calculate each segment length

The total length \(6\sqrt{13}\) must be split into three equal segments. Therefore, each segment's length should be \(\frac{6\sqrt{13}}{3} = 2\sqrt{13}\).

Step 3: Verify the distance from (3,6) to (7,12)

Now we calculate the distance from the starting point \((3,6)\) to the first dividing point \((7,12)\) using the distance formula:\[ d_1 = \sqrt{(7 - 3)^2 + (12 - 6)^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \]

Step 4: Calculate the distance from (7,12) to (11,18)

Next, evaluate the distance between the two dividing points \((7,12)\) and \((11,18)\):\[ d_2 = \sqrt{(11 - 7)^2 + (18 - 12)^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \]

Step 5: Verify distance from (11,18) to (15,24)

Finally, compute the distance from the second dividing point \((11,18)\) to the endpoint \((15,24)\):\[ d_3 = \sqrt{(15 - 11)^2 + (24 - 18)^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \]

Step 6: Conclusion

All three distances \(d_1\), \(d_2\), and \(d_3\) are equal to \(2\sqrt{13}\). This confirms that the points \((7,12)\) and \((11,18)\) divide the line segment into three equal parts, verifying the original statement.

Key Concepts

Coordinate Geometry

Coordinate geometry is the study of geometric figures through a coordinate system. It allows us to use algebraic methods for solving geometric problems. In coordinate geometry, points are placed on a plane using ordered pairs

• The first number in the pair represents the x-coordinate, which measures the horizontal distance from the origin.
• The second number represents the y-coordinate, which indicates the vertical distance from the origin.

Understanding the placement of points on a coordinate plane is important for calculating distances. We often use the distance formula, which stems from the Pythagorean theorem, to determine the distance between two points. The standard formula for distance \( d \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

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

Applying this formula helps in dividing line segments and performing other geometric operations on a coordinate plane.

Equal Segments

Equal segments mean that a line segment is divided into smaller parts of equal length. For example, if a line is divided into three equal segments, then each part shares the same measurement. This concept is essential in coordinate geometry, as you can use it to confirm that points divide a segment evenly.
In our example:

• We started with endpoints \( (3,6) \) and \( (15,24) \), finding the total distance using the formula.
• The total distance was divided into three, as required, to determine the length of each equal part.
• Each division had a length of \( 2\sqrt{13} \), ensuring the division into equal segments.

Checking each part involved recalculating the distance from each dividing point to confirm it matched the calculated segment length. Once equal lengths are confirmed, it ensures the points effectively divide the line into three identical fractions.

Intermediate Algebra

Intermediate Algebra involves working with equations and simplifying expressions, but it also plays a crucial role in coordinate geometry. The field introduces concepts such as functions, graphing, and linear equations.
In this problem:

• The distance formula is not just arithmetic; it involves algebraic manipulation, requiring square rooting and understanding relations between coordinates.
• We must simplify expressions to find lengths and compare segments.

Leveraging intermediate algebra skills supports tasks like:

• Streamlining calculations
• Understanding the geometric representation of algebraic expressions

It bridges the gap between pure algebra and its applications, helping to solve real-world geometric problems through calculation and analysis.