Problem 4 - Algebra and Trigonometry

Question

\(1-6=\) Show that the point is on the unit circle. $$ \left(-\frac{5}{7},-\frac{2 \sqrt{6}}{7}\right) $$

Step-by-Step Solution

Answer

Yes, the point is on the unit circle because \(x^2 + y^2 = 1\).

Step 1: Understand the unit circle definition

A point is on the unit circle if the sum of the squares of its coordinates equals 1. This means for a point \((x, y)\), it must satisfy the equation \(x^2 + y^2 = 1\).

Step 2: Substitute the coordinates

Given the point \(\left(-\frac{5}{7}, -\frac{2\sqrt{6}}{7}\right)\), substitute \(x = -\frac{5}{7}\) and \(y = -\frac{2\sqrt{6}}{7}\) into the equation \(x^2 + y^2 = 1\).

Step 3: Calculate \(x^2\)

Find \(x^2\) by squaring \(-\frac{5}{7}\): \[x^2 = \left(-\frac{5}{7}\right)^2 = \frac{25}{49}\.\]

Step 4: Calculate \(y^2\)

Find \(y^2\) by squaring \(-\frac{2\sqrt{6}}{7}\):\[y^2 = \left(-\frac{2\sqrt{6}}{7}\right)^2 = \frac{4 \times 6}{49} = \frac{24}{49}.\]

Step 5: Add \(x^2\) and \(y^2\)

Add the calculated values of \(x^2\) and \(y^2\):\[x^2 + y^2 = \frac{25}{49} + \frac{24}{49} = \frac{49}{49} = 1.\]

Step 6: Conclusion

Since \(x^2 + y^2 = 1\), the point \(\left(-\frac{5}{7}, -\frac{2\sqrt{6}}{7}\right)\) is on the unit circle.

Key Concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is a branch of geometry where we describe and analyze geometric shapes using a coordinate system. In the case of the unit circle, we use the Cartesian coordinate system, involving an x-axis and a y-axis, to locate points. A unit circle is a special geometric figure with its center at the origin (0, 0) and a radius of 1 unit.
The primary equation used to represent the unit circle in coordinate geometry is:

• \(x^2 + y^2 = 1\)

This equation implies that any point (x, y) located on the unit circle must satisfy the condition that the sum of squares of its x-coordinate and y-coordinate equals 1.
When analyzing points, it is essential to plug them into this equation to verify their position on the unit circle. By substituting the coordinates into the equation, you can determine whether the point is indeed on the circle, offering a clear intersection between algebra and geometry through coordinate-based analysis.

Circle Equation

The circle equation is a mathematical expression that describes all the points located at a specific distance, termed as the radius, from a fixed point called the center. For the unit circle:

• The center is at (0, 0).
• The radius is 1.

This simplifies the general circle equation

• \((x - h)^2 + (y - k)^2 = r^2\)

to

• \(x^2 + y^2 = 1\)

for the unit circle.
To show that a point lies on the unit circle, substitute the x and y coordinates of the point into the equation. For example, to confirm the position of the point \((-\frac{5}{7}, -\frac{2\sqrt{6}}{7})\), you calculate:

• \(x^2 = \left(-\frac{5}{7}\right)^2\).
• \(y^2 = \left(-\frac{2\sqrt{6}}{7}\right)^2\).

Adding these values, if they equal 1, confirms the point is on the circle. This method uses algebraic manipulation to verify relationships in geometric contexts.

Algebraic Proof

An algebraic proof uses a sequence of logical algebraic steps to demonstrate a truth. In the context of the unit circle, such a proof shows that given points satisfy the circle's defining equation. You begin by taking the known coordinates of the point, such as \((-\frac{5}{7}, -\frac{2\sqrt{6}}{7})\), and calculating their respective squares.
For this point, calculate:

• \(x^2 = \frac{25}{49}\)
• \(y^2 = \frac{24}{49}\)

Then, add these squared values to check their sum:

• \(x^2 + y^2 = \frac{25}{49} + \frac{24}{49} = \frac{49}{49} = 1\)

This verification through calculation shows the point fulfills the equation \(x^2 + y^2 = 1\), hence lies on the unit circle.
Algebraic proofs are powerful as they provide clarity and certainty in mathematical reasoning. By ensuring each step logically follows the previous one, students can understand how abstract concepts are validated mathematically.