Introduction

In the realm of geometry, proving that a quadrilateral is cyclic (i.e., can be inscribed in a circle) can be a challenging yet fascinating task. For the given points A(0, 0), B(3, 1), C(1, 2), and D(2, -1), we will demonstrate how to prove that the quadrilateral formed by these points is cyclic using geometric and algebraic methods.

Geometric Method with Slope

To prove that a quadrilateral is cyclic, one of the most straightforward methods is to show that the opposite angles are supplementary (i.e., add up to 180 degrees). However, an alternative approach involves using the slopes of the sides and diagonals. Let's proceed step-by-step.

Step 1: Calculate Slopes

We start by calculating the slopes of all sides and diagonals of the quadrilateral ABCD.

Calculating Slopes of Sides

Slope of AC:

Slope formula: (y2 - y1) / (x2 - x1)

Calculation: Slope AC (2 - 0) / (1 - 0) 2

Slope of CB:

Calculation: Slope CB (1 - 2) / (3 - 1) -1/2

Slope of BD:

Calculation: Slope BD (-1 - 1) / (2 - 3) -2/-1 2

Slope of DA:

Calculation: Slope DA (-1 - 0) / (2 - 0) -1/2

Step 2: Analyze Opposite Angles

The next step is to determine if the opposite angles are supplementary.

Consider Angles:

Angle ACB: Since the slope of AC is 2 and the slope of CB is -1/2, the product of the slopes is 2 * (-1/2) -1. This indicates that the lines are perpendicular, meaning angle ACB is 90 degrees.

Angle CBD: Similarly, the slope of CB is -1/2 and the slope of BD is 2, and their product is also -1, indicating that angle CBD is 90 degrees.

Angle BDA: The slope of BD is 2 and the slope of DA is -1/2. Again, their product is -1, making angle BDA 90 degrees.

Angle DAC: Conversely, the slope of DA is -1/2 and the slope of AC is 2. Their product is -1, making angle DAC 90 degrees.

Since all four angles are 90 degrees, the sum of opposite angles (ACB BDA and CBD DAC) is 180 degrees, proving that quadrilateral ABCD is cyclic.

Algebraic Method with Perpendicularity and Equal Magnitude

Another method to prove that a quadrilateral is cyclic is to show that the diagonals are perpendicular and have equal magnitude. Additionally, the sides of the quadrilateral can also be perpendicular and of equal magnitude.

Step 1: Check Perpendicularity of Diagonals

Let's check the diagonals AC and BD.

Diagonal AC:

Length AC √[(3-0)2 (1-0)2] √(9 1) √10

Diagonal BD:

Length BD √[(2-3)2 (-1-1)2] √(1 4) √5

Since the diagonals do not have the same magnitude, we need to check for perpendicularity.

Diagonal AC and BD:

Slope of AC 2, Slope of BD 2 (from our previous calculations)

Product of slopes 2 * 2 4, which is not -1 (indicating they are not perpendicular).

Step 2: Check for Perpendicularity and Equal Magnitude of Sides

Now, let's check the sides AC and AD, as well as AB and CD.

Sides AC and AD:

Slope of AC 2, Slope of AD -1/2

Product of slopes 2 * (-1/2) -1 (indicating they are perpendicular and of equal magnitude).

Length AC √(10) and Length AD √(5)

Sides AB and CD:

Slope of AB (1-0) / (3-0) 1/3, Slope of CD (2--1) / (1-2) 3/-1 -3

Product of slopes (1/3) * (-3) -1 (indicating they are perpendicular and of equal magnitude).

Length AB √(10) and Length CD √(10)

Conclusion

Given the above calculations, we can conclude that the quadrilateral ABCD can be classified as a square. Squares are cyclic quadrilaterals, as all their angles are 90 degrees and all their sides are equal. Therefore, quadrilateral ABCD is indeed cyclic.

In summary, by using both the geometric approach with slopes and the algebraic approach with perpendicularity and equal magnitude, we have proven that the quadrilateral formed by points A(0, 0), B(3, 1), C(1, 2), and D(2, -1) is cyclic.