Problem 12 - Algebra and Trigonometry

Question

\(9-14\) . Find the missing coordinate of \(P\) , using the fact that \(P\) lies on the unit circle in the given quadrant. $$ \frac{\text { Coordinates }}{P\left(\frac{2}{5},\right.} ) \frac{\text { Quadrant }}{\text { I }} $$

Step-by-Step Solution

Answer

The missing coordinate is \( \frac{\sqrt{21}}{5} \), so \( P = \left(\frac{2}{5}, \frac{\sqrt{21}}{5}\right) \).

Step 1: Understanding the Unit Circle

A point on the unit circle is a point \( P(x, y) \) such that \( x^2 + y^2 = 1 \). This comes from the equation of the unit circle, where the radius is 1.

Step 2: Substitute Known Coordinate

Substitute the known x-coordinate into the unit circle equation: \(\left(\frac{2}{5}\right)^2 + y^2 = 1\).

Step 3: Solve for y

Compute \( \left(\frac{2}{5}\right)^2 = \frac{4}{25} \). Then substitute this into the equation: \( \frac{4}{25} + y^2 = 1 \). Subtract to find \( y^2 = 1 - \frac{4}{25} \).

Step 4: Simplify Equation

Simplify \( 1 - \frac{4}{25} = \frac{25}{25} - \frac{4}{25} = \frac{21}{25} \). So, \( y^2 = \frac{21}{25} \).

Step 5: Solve for y using Square Roots

Take the square root of both sides: \( y = \pm \sqrt{\frac{21}{25}} \). Since \( P \) is in the first quadrant, both coordinates must be positive. Therefore, \( y = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5} \).

Step 6: Conclusion

The full coordinates for \( P \) are \( \left(\frac{2}{5}, \frac{\sqrt{21}}{5}\right) \).

Key Concepts

Quadrants

In coordinate geometry, the Cartesian plane is divided into four distinct areas known as quadrants. These quadrants are defined by the x-axis and y-axis intersecting at right angles, creating four sections. Each quadrant is assigned a number from I to IV, usually written in Roman numerals, and they are ordered in a counterclockwise direction starting from the upper right quadrant.

- **Quadrant I**: Here, both the x and y coordinates of any point are positive. This is where our point of interest, P, is located. - **Quadrant II**: In this section, the x-coordinate is negative while the y-coordinate remains positive. - **Quadrant III**: Both the x and y coordinates are negative in this quadrant. - **Quadrant IV**: The x-coordinate is positive, but the y-coordinate is negative.

Understanding which quadrant a point lies in helps in determining the sign of its coordinates, which is crucial when solving problems on the unit circle.

Equation of a Circle

The equation of a circle in standard form is given by \[ (x - h)^2 + (y - k)^2 = r^2 \]where \((h, k)\) is the center of the circle and \(r\) is its radius. However, when dealing with a unit circle, the equation simplifies significantly because the radius is always 1, and the circle is centered at the origin \((0, 0)\). The equation then becomes:\[ x^2 + y^2 = 1 \]

This equation tells us that any point \((x, y)\) lying on the unit circle will satisfy this equation. The simple equality highlights the geometric fact that the distance from any point on the circle to the origin is exactly 1. This concept is essential in solving problems involving the unit circle, such as finding missing coordinates, as demonstrated with point P in the solution.

Coordinate Geometry

Coordinate geometry, or analytic geometry, involves the study of geometric figures using a coordinate system. By representing geometric shapes and figures with algebraic equations, problems in plane geometry become easier to address with mathematics. One fundamental aspect of coordinate geometry is using the Cartesian plane to describe the location of points by assigning them coordinates \((x, y)\).

In the context of the unit circle problem, coordinate geometry helps by converting a visual geometric problem into an algebraic one. By knowing the x-coordinate of a point \(P\) and using the unit circle equation \(x^2 + y^2 = 1\), we can algebraically solve for the y-coordinate. This showcases the power and utility of coordinate geometry — transforming a visual geometric question into an algebraic equation that can be solved systematically. This approach not only aids in gaining deeper insights but also ensures precise solutions in geometrical contexts.