Problem 11 - 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. $$ \frac{\text { Coordinates }}{P\left(\frac{1}{3},\right.} ) \frac{\text { Quadrant }}{\text { II }} $$

Step-by-Step Solution

Answer

The coordinate \(y\) is \(\frac{2\sqrt{2}}{3}\), as point \(P\) is in Quadrant II.

Step 1: Understand the Unit Circle

The unit circle is a circle with a radius of 1 centered at the origin, (0,0), in the coordinate plane. Every point \((x, y)\) on this circle satisfies the equation \(x^2 + y^2 = 1\). The reason for this is that the radius is the constant 1, so by the Pythagorean theorem, the sum of the squares of the x- and y-coordinates must equal 1.

Step 2: Substitute the Known Coordinate

We are given that \(P\left(\frac{1}{3}, y\right)\) lies on the unit circle. Substitute \(x = \frac{1}{3}\) into the equation of the unit circle: \(\left(\frac{1}{3}\right)^2 + y^2 = 1\).

Step 3: Simplify and Solve for \(y^2\)

Calculate \(\left(\frac{1}{3}\right)^2 = \frac{1}{9}\). Substitute it back in: \(\frac{1}{9} + y^2 = 1\). Move \(\frac{1}{9}\) to the other side to get: \(y^2 = 1 - \frac{1}{9}\).

Step 4: Simplify the Right-Hand Side

Calculate \(1 - \frac{1}{9}\). This equals \(\frac{9}{9} - \frac{1}{9} = \frac{8}{9}\). So, \(y^2 = \frac{8}{9}\).

Step 5: Solve for \(y\)

Take the square root of both sides: \(y = \pm \sqrt{\frac{8}{9}}\). This simplifies to \(y = \pm \frac{\sqrt{8}}{3}\). Further simplify \(\sqrt{8}\) as \(2\sqrt{2}\), so \(y = \pm \frac{2\sqrt{2}}{3}\).

Step 6: Determine the Sign of \(y\) in Quadrant II

In Quadrant II, the x-coordinates are negative, and the y-coordinates are positive. Since \(x = \frac{1}{3}\) is positive, and we have already squared it, we don't need to change signs here. But \(y\) must be positive, so we select \(y = \frac{2\sqrt{2}}{3}\).

Key Concepts

Quadrants in coordinate geometry

In coordinate geometry, a plane is divided into four regions called quadrants. These quadrants help us understand the position of a point relative to the origin. Each quadrant corresponds to a different combination of positive and negative signs for the x and y coordinates.

Here are the characteristics of each quadrant:

• **Quadrant I:** Both x and y coordinates are positive.
• **Quadrant II:** The x coordinate is negative, and the y coordinate is positive.
• **Quadrant III:** Both x and y coordinates are negative.
• **Quadrant IV:** The x coordinate is positive, and the y coordinate is negative.

In the exercise, point P is located in Quadrant II, which is why its x coordinate is negative, while its y coordinate is positive. This is crucial to correctly deducing the value of y once solved.

Equation of a circle

The equation of a circle is a fundamental concept in geometry that describes all the points lying on a circle. A circle is defined by its center and radius.

For a circle centered at the origin and with radius 1, the equation becomes:
\[ x^2 + y^2 = r^2 \rightarrow x^2 + y^2 = 1 \]
This specific form is known as the equation of the unit circle. Any point \((x, y)\) that satisfies this equation lies on the unit circle. Knowing this makes solving the problem easier, as we start with this equation to find the missing coordinate of point P.

Importantly, the subtraction of squares and how they equal 1 connects deeply with the Pythagorean Theorem, which is versatile in geometric calculations.

Pythagorean theorem

The Pythagorean theorem is a cornerstone in both geometry and trigonometry, stating that for a right-angled triangle, the square of the hypotenuse's length is the sum of the squares of the other two sides.

Expressed mathematically, it is written as:
\[ a^2 + b^2 = c^2 \]
Here, \(c\) is the hypotenuse. In a unit circle, the hypotenuse is always 1, so the Pythagorean theorem simplifies to the equation of the circle formula:
\[ x^2 + y^2 = 1 \]
We used this relationship in solving for the missing coordinate of the point on the unit circle. With a known x value, the theorem helps ensure consistency with our y calculation, by confirming it satisfies the circle's equation. This is essentially what determines the missing y value when x is given.