Problem 42 - Precalculus Mathematics for Calculus

Question

Show that the points \(A(-1,3), B(3,11),\) and \(C(5,15)\) are collinear by showing that \(d(A, B)+d(B, C)=d(A, C)\)

Step-by-Step Solution

Answer

Points A, B, and C are collinear since \(d(A, B) + d(B, C) = d(A, C)\).

Step 1: Calculate Distance AB

To calculate the distance between points A(-1,3) and B(3,11), use the distance formula: \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]Substitute in the values from points A and B: \[d(A, B) = \sqrt{(3 - (-1))^2 + (11 - 3)^2} = \sqrt{(3 + 1)^2 + (8)^2} = \sqrt{4^2 + 8^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}\]

Step 2: Calculate Distance BC

Next, find the distance between points B(3,11) and C(5,15) using the distance formula: \[d(B, C) = \sqrt{(5 - 3)^2 + (15 - 11)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}\]

Step 3: Calculate Distance AC

Now, compute the distance between points A(-1,3) and C(5,15):\[d(A, C) = \sqrt{(5 - (-1))^2 + (15 - 3)^2} = \sqrt{(5 + 1)^2 + 12^2} = \sqrt{6^2 + 12^2} = \sqrt{36 + 144} = \sqrt{180} = 6\sqrt{5}\]

Step 4: Verify Collinearity

To verify that the points are collinear, check if the sum of distances AB and BC equals the distance AC:\[d(A, B) + d(B, C) = 4\sqrt{5} + 2\sqrt{5} = 6\sqrt{5}\]Since this is equal to \(d(A, C) = 6\sqrt{5}\), points A, B, and C are collinear.

Key Concepts

Distance Formula

The distance formula is a fundamental tool in coordinate geometry that helps in calculating the distance between two points in a plane. It's derived from the Pythagorean theorem and is expressed as:\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]This formula applies to any two points \(x_1, y_1\) and \(x_2, y_2\) on a coordinate plane. To find the distance between these points, you first calculate the difference in the x-coordinates and the y-coordinates, then square these differences.

• Add the squared differences together
• Finally, take the square root of this sum to get the distance

Using the distance formula allows you to easily confirm the collinearity of points or solve various problems in geometry that require precise measurements between points.

Coordinate Geometry

Coordinate geometry, often referred to as analytic geometry, combines the principles of algebra and geometry using the coordinate plane. It is a powerful area of mathematics that enables solving geometric problems numerically. In this field, points are expressed as ordered pairs \(x, y\) showing their exact location on a plane.
With coordinate geometry, you can:

• Determine the distance and midpoint between points
• Find slopes and equations of lines
• Establish the relationships between shapes and their equations

This integration of algebra with geometry makes it easier to prove concepts such as collinearity, as seen in the original exercise where the distances between points show they lie on the same straight line. Geometry comes to life when mapped on a coordinate plane, providing a visual understanding while solving algebraic problems.

Precalculus Mathematics

Precalculus mathematics acts as a bridge between algebra and calculus, building a foundation for advanced mathematical concepts. It is essential for students to grasp introductory topics such as functions, complex numbers, and coordinate geometry before moving to calculus.
In the context of the given problem, precalculus provides a toolkit that includes the distance formula, enabling students to solve problems analytically.

• It nurtures logical reasoning and problem-solving skills
• Prepares students for calculus by practicing essential concepts
• Encourages comprehension of geometric properties and their implications

Mastering these basics in precalculus mathematics not only helps in current problem-solving but also paves the way for tackling more complex calculus challenges involving limits, derivatives, and integrals.