Problem 6 - Algebra and Trigonometry

Question

\(1-6=\) Show that the point is on the unit circle. $$ \left(\frac{\sqrt{11}}{6}, \frac{5}{6}\right) $$

Step-by-Step Solution

Answer

The point \(\left(\frac{\sqrt{11}}{6}, \frac{5}{6}\right)\) is on the unit circle.

Step 1: Understand the Unit Circle Equation

The equation of a unit circle centered at the origin is \(x^2 + y^2 = 1\). To determine if a point \((x, y)\) lies on the unit circle, substitute the coordinates into this equation and check whether the equation holds true.

Step 2: Substitute the Point's Coordinates

Given the point \(\left(\frac{\sqrt{11}}{6}, \frac{5}{6}\right)\), substitute \(x = \frac{\sqrt{11}}{6}\) and \(y = \frac{5}{6}\) into the unit circle equation: \(\left(\frac{\sqrt{11}}{6}\right)^2 + \left(\frac{5}{6}\right)^2 = 1\).

Step 3: Calculate the Squares

Calculate \(\left(\frac{\sqrt{11}}{6}\right)^2 = \frac{11}{36}\) and \(\left(\frac{5}{6}\right)^2 = \frac{25}{36}\).

Step 4: Add the Squares

Add the results: \(\frac{11}{36} + \frac{25}{36} = \frac{36}{36} = 1\).

Step 5: Verify the Result

Since \(\frac{36}{36} = 1\), the point \(\left(\frac{\sqrt{11}}{6}, \frac{5}{6}\right)\) satisfies the unit circle equation \(x^2 + y^2 = 1\). Thus, the point is on the unit circle.

Key Concepts

Coordinate Geometry

Coordinate geometry, often referred to as analytic geometry, is a branch of mathematics that studies geometric objects using a coordinate system. This integration of algebra and geometry allows us to examine various shapes and figures by representing them with equations.
One of the simplest forms in coordinate geometry is the point, which is defined by a pair of coordinates

• The first number in the pair specifies its horizontal position (x-coordinate).
• The second number specifies its vertical position (y-coordinate).

To better understand, imagine placing a dot on a piece of graph paper. The exact position of the dot can be described by how far it is along the x-axis and the y-axis. This comprehensive mapping makes coordinate geometry vital for precisely describing locations in a two-dimensional space, like demonstrating that a point lies on a circle or line.

Equation of a Circle

In coordinate geometry, circles can be represented through specific equations. The standard equation of a circle is derived from its definition: a set of all points equidistant from a central point, known as the center.
For a circle centered at the origin

• The general equation is given by \[ x^2 + y^2 = r^2 \]
• Where \( r \) is the radius of the circle.

In the special case of a unit circle, where the radius is exactly 1, the equation simplifies to \[ x^2 + y^2 = 1 \]. This form allows easy verification of points on the circle by substituting their coordinates into the equation, which is a common exercise to test understanding of circular equations in coordinate geometry.

Verification of a Point on a Circle

Verifying whether a point lies on a circle involves a straightforward substitution process using the circle's equation. This process is essential in coordinate geometry for confirming theoretical results.
To verify a point on a unit circle:

• Start with the circle's equation \[ x^2 + y^2 = 1 \]
• Insert the x and y values from the point’s coordinates into this equation.

As detailed in the exercise, given a point \( \left(\frac{\sqrt{11}}{6}, \frac{5}{6}\right) \), the substitution involves calculating \[ \left(\frac{\sqrt{11}}{6}\right)^2 + \left(\frac{5}{6}\right)^2 \]. Simplifying this gives the equation \[ \frac{11}{36} + \frac{25}{36} = \frac{36}{36} = 1 \].The conclusion is that the point indeed lies on the unit circle, as it satisfies the equation. This method provides a practical application of understanding and using circle equations in geometry, reinforcing the fundamental connection between algebra and spatial reasoning.